Growth theory of subharmonic functions

I ••

11

Birkhäuser Advanced Texts Edited by Herbert Amann, Zürich University Steven G. Krantz, Washington University, St. Louis Shrawan Kumar, University of North Carolina at Chapel Hill Jan Nekovár, ˇ Université Pierre et Marie Curie, Paris

Vladimir Azarin

Growth Theory of Subharmonic Functions

Birkhäuser Basel · Boston · Berlin

Author: Vladimir Azarin Department of Mathematics Bar-Ilan University Ramat-Gan 52900 Israel e-mail: [email protected]

2000 Mathematics Subject Classification: Primary 31B05; Secondary 30D20

Library of Congress Control Number: 2008930978

Bibliographic information published by Die Deutsche Bibliothek Die Deutsche Bibliothek lists this publication in the Deutsche Nationalbibliografie; detailed bibliographic data is available in the Internet at .

ISBN 978-3-7643-8885-0 Birkhäuser Verlag AG, Basel · Boston · Berlin This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, re-use of illustrations, recitation, broadcasting, reproduction on microfilms or in other ways, and storage in data banks. For any kind of use permission of the copyright owner must be obtained. © 2009 Birkhäuser Verlag AG Basel · Boston · Berlin P.O. Box 133, CH-4010 Basel, Switzerland Part of Springer Science+Business Media Printed on acid-free paper produced of chlorine-free pulp. TCF ∞ Printed in Germany ISBN 978-3-7643-8885-0

e-ISBN 978-3-7643-8886-7

987654321

www.birkhauser.ch

Contents 1

Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2 Auxiliary Information. Subharmonic Functions 2.1 Semicontinuous functions . . . . . . . . . . . . . . . . . . . 2.2 Measures and integrals . . . . . . . . . . . . . . . . . . . . . 2.3 Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Harmonic functions . . . . . . . . . . . . . . . . . . . . . . . 2.5 Potentials and capacities . . . . . . . . . . . . . . . . . . . . 2.6 Subharmonic functions . . . . . . . . . . . . . . . . . . . . . 2.7 Sequences of subharmonic functions . . . . . . . . . . . . . 2.8 Scale of growth. Growth characteristics of subharmonic functions . . . . . . . . . . . . . . . . . . . 2.9 The representation theorem of subharmonic functions in Rm

. . . . . . .

3 7 14 21 28 36 43

. . . . . . . .

52 62

3 Asymptotic Behavior of Subharmonic Functions of Finite Order 3.1 Limit sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Indicators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Densities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

75 87 99

4 Structure of Limit Sets 4.1 Dynamical systems . . . . . . . . . . . . . . . . . . . . 4.2 Subharmonic function with prescribed limit set . . . . 4.3 Further properties of limit sets . . . . . . . . . . . . . 4.4 Subharmonic curves. Curves with prescribed limit sets

. . . .

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. . . .

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1

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. . . .

5 Applications to Entire Functions 5.1 Growth characteristics of entire functions . . . . . . . . . . . . . . 5.2 D -topology and topology of exceptional sets . . . . . . . . . . . . 5.3 Asymptotic approximation of subharmonic functions . . . . . . . . 5.4 Lower indicator of A.A. Gol dberg. Description of lower indicator. Description of the pair: indicator-lower indicator . . . . . . . . . . 5.5 Asymptotic extremal problems. Semiadditive integral . . . . . . . .

107 121 134 147 151 152 157 163 173

vi

Contents

5.6 5.7 5.8

Entire functions of completely regular growth. Levin-Pfluger Theorem. Balashov’s theory . . . . . . . . . . . . . . 177 General characteristics of growth of entire functions . . . . . . . . 180 A generalization of the Valiron-Titchmarsh theorem . . . . . . . . 199

6 Application to the Completeness of Exponential Systems 6.1 The multiplicator problem . . . . . . . . . . . . . . . . . . . . . . . 203 6.2 A generalization of ρ-trigonometric convexity . . . . . . . . . . . . 223 6.3 Completeness of exponential systems in convex domains . . . . . . 228 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247 List of Terms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 249 Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255

Chapter 1

Preface This book aims to convert the noble art of constructing an entire function with prescribed asymptotic behavior to a handicraft. For this you should just consider the limit set that describes the asymptotic behavior of the entire function, i.e., you should consider the set U [ρ, σ] of subharmonic functions (that is, {v is subharmonic : v(reiφ ) ≤ σrρ }) and pick out the subset U which characterizes its asymptotic properties. How to do it? The properties of limit sets are listed in Section 3. All the standard growth characteristics are expressed in terms of limit sets in Sections 3.2, 3.3, 5.7. Examples of construction are to be found in Sections 5.4–6.3. So you can use this book as a reference book for construction of entire functions. Of course, you need some terms. All the terms that we use in this book are listed on pages 249–253. If you want to study the theory, I recommend that you solve the exercises that are in the text. Most of them are trivial. However, I recommend that you do all of them by the moment that they appear trivial to you. A few words about the history of this book. It arose from a course of lectures that I gave at Kharkov University in 1977. After some time, under pressure and with active help of Prof. I.V. Ostrovskii, a rotaprint edition (Edition of KhGU)of this course appeared: the first part in 1978, the second one in 1982. Mathematical Reviews did not notice this fact. Since that time lots of new and important results have been obtained. Some of them were presented in Chapter 3 of the review [GLO]. In 1994, when I started to work in the Bar-Ilan University and obtained a personal computer, my first wish was to study typing on it in English. This was the first impulse for translating this course into English (there are no more than five copies of this book in the world, I believe, one of them being mine). I continued this project while working in Bar-Ilan (1994–2006) but there was not much time for this. And now I have finished.

2

Chapter 1. Preface

Acknowledgements I am indebted to many people. I start from Prof. I.V. Ostrovskii, who supported this idea for many years, and Prof. A.A. Gol dberg, who stimulated my mathematical activity all my life by his letters and conversations. I am indebted to Prof’s A. Eremenko and M. Sodin, who, not being my “aspirants,” solved a lot of problems connected to limit sets, and also to Dr.’s V. Giner, L. Podoshev and E. Fainberg who worked with me to develop the theory. I am indebted to Prof’s L. H¨ ormander and R. Sigurdsson who have sent me the preprints of their papers that were not yet published. I am indebted to Prof. I.F. Krasichkov–Ternovskii, who explained to me many years ago the connection between the multiplicator problem and completeness of the exponent system in a convex domain. I am indebted to Prof’s. M.I. Kadec and V.P. Fonf for proving Theorem 4.1.5.2, which is rather far off my speciality. I am indebted to my coauthors Prof’s D. Drasin and P. Poggi–Corradini; I have exploited the results of our joint paper in Section 6.2. Of course, I am indebted to my late teacher Prof. B.Ya. Levin, who taught me entire and subharmonic functions and gave me the first problems in this area. Actually, the theory of limit sets is a generalization of the theory of functions of completely regular growth. I am also indebted very much to my grandson Sasha Sodin, who transformed “my English” into English.

Chapter 2

Auxiliary Information. Subharmonic Functions 2.1 Semicontinuous functions 2.1.1. Let x ∈ Rm be a point in an m-dimensional Euclidean space, E a Borel set and f (x) a function on E such that f (x) = ∞. Set (2.1.1.1) M (f, x, ε) := sup{f (x ) : |x − x | < ε, x ∈ E} The function

f ∗ (x) := lim M (f, x, ε) ε→0

is called the upper semicontinuous regularization of the function f (x). In the case of a finite jump, the regularization “raises” the values of the function. However, there is no influence on f ∗ (x), if f (x) tends to −∞ “continuously”. Proposition 2.1.1.1 (Regularization Properties) The following properties hold: (rg1)

f (x) ≤ f ∗ (x);

(rg2)

(αf )∗ (x) = αf ∗ (x);

(rg3)

(f ∗ )∗ (x) = f ∗ (x);

(rg4)

(f1 + f2 )∗ (x) ≤ f1∗ (x) + f2∗ (x); (max(f1 , f2 ))∗ (x) ≤ max(f1∗ , f2∗ )(x); (min(f1 , f2 ))∗ (x) ≤ min(f1∗ , f2∗ )(x). These properties are obvious corollaries of the definition of f ∗ (x).

4

Chapter 2. Auxiliary Information. Subharmonic Functions

Exercise 2.1.1.1 Prove them. 2.1.2 The function f (x) is called upper semicontinuous at a point x if f ∗ (x) = f (x). We denote the class of upper semicontinuous functions on E by C + (E). The function f (x) is called lower semicontinuous if −f (x) is upper semicontinuous (notation f ∈ C − (E)). Examples of semicontinuous functions are given by Proposition 2.1.2.1 (Semicontinuity of Characteristic Functions of Sets) Let G ⊂ Rm be an open set. Then its characteristic function χG is lower semicontinuous in Rm . Let F be a closed set, then χF is upper semicontinuous. The proof is obvious. Exercise 2.1.2.1 Prove this. Proposition 2.1.2.2 (Connection with Continuity) If f ∈ C + ∩ C − , then f is continuous. The assertion follows from the equalities f ∗ (x) = lim sup{f (x ) : |x − x | < ε}; −(−f )∗ (x) = lim inf {f (x ) : |x − x | < ε}. ε→0

ε→0

Proposition 2.1.2.3 (C + -Properties) The following holds: (C + 1)

f ∈ C + (E) ⇒ αf ∈ C + (E), for α ≥ 0

(C + 2)

f1 , f2 ∈ C + ⇒ f1 + f2 , max(f1 , f2 ), min(f1 , f2 ) ∈ C + .

These properties follow from the properties of regularization (Proposition 2.1.1.1). Exercise 2.1.2.2 Prove them. Let G be an open set. Set GA := {x ∈ G : f (x) < A}. Theorem 2.1.2.4 (First Criterion of Semicontinuity) One has f ∈ C + if and only if GA is open for all A ∈ R. Proof. Let f (x) = f ∗ (x), x ∈ G. Then {f (x) < A} =⇒ {f ∗ (x) < A} =⇒ {M (f, x, ε) < A} for all sufficiently small ε. Thus the neighborhood of x Vε,x := {x : |x − x | < ε} is contained in GA . Conversely, since the set GA is open for A = f (x0 ) + δ, we have f ∗ (x0 ) ≤ f (x0 )+ δ for any δ > 0, hence for δ = 0. With property (rg1) of Proposition 2.1.1.1  this gives f ∗ (x0 ) = f (x0 ). Let F be a closed set. Set F A := {x ∈ F : f (x) ≥ A}. An obvious corollary of the previous theorem is

2.1. Semicontinuous functions

5

Corollary 2.1.2.5 One has f ∈ C + if and only if F A is closed for all A. Exercise 2.1.2.3 Prove the corollary. We denote compact sets by K. Set M (f, K) = sup{f (x) : x ∈ K}. Theorem 2.1.2.6 (Weierstrass) Let K ⊂ Rm be a compact set and f ∈ C + (K). Then there exists x0 ∈ Ksuch that f (x0 ) = M (f, K). I.e., f attains its supremum on any compact set. Proof. Set Kn := {x ∈ K : f (x) ≥ M (f, K) − 1/n}. The Kn are closed by Corollary 2.1.2.5, nonempty by definition of M (f, K). Their intersection is nonempty and is equal to the set Kmax := {x ∈ K : f (x) ≥ M (f, K)}. It means that there exists x0 in K such that f (x0 ) ≥ M (f, K). The opposite inequality holds for any x in K.



Exercise 2.1.2.4 Why? The following theorem shows that the functional M (f, K) is continuous with respect to monotonic convergence of semicontinuous functions. Proposition 2.1.2.7 (Continuity from the right of M (f, K)) Let fn ∈ C + (K), fn ↓ f, n = 1, 2, 3 . . . . Then M (fn , K) ↓ M (f, K). Proof. It is clear that lim M (fn , K) := M exists. n→∞

of the closed nonempty Set Kn := {x ∈ K : fn (x) ≥ M }. The intersection  sets Kn is nonempty and has the following form: Kn = {x : f (x) ≥ M }. So n

M (f, K) ≥ M . The opposite inequality is obvious.



Exercise 2.1.2.5 Why? In the same way one proves Proposition 2.1.2.8 (Commutativity of inf and M (·)) Let {fα ∈ C+ (K), α ∈ (0; ∞)} be an arbitrarily decreasing family of semicontinuous functions. Then inf M (fα , K) = M (inf fα , K). α

α

6

Chapter 2. Auxiliary Information. Subharmonic Functions

Exercise 2.1.2.6 Prove this proposition. Theorem 2.1.2.9 (Second Criterion of Semicontinuity) f ∈ C + (K) iff there exists a sequence fn of continuous functions such that fn ↓ f . Sufficiency. Let fn ∈ C + (K), fn ↓ f .Set KnA := {x ∈ K : fn (x) ≥ A}. This is a sequence of nonempty closed If the set K A := {x : f (x) ≥ A} is nonempty,  sets. A A then K is closed because Kn = K A . Hence f ∈ C + (K) by Corollary 2.1.2.5. n

Necessity. Set fn (x, y) := f (y) − n|x − y|. This sequence of functions has the following properties: a) it decreases monotonically in n and  lim fn (x, y) =

n→∞

f (x), for x = y; −∞, for x = y;

b) for any fixed n the functions fn are continuous in x uniformly with respect to y, because |fn (x, y) − fn (x , y)| ≤ n|x − x |; c) fn are upper semicontinuous in y. Proposition 2.1.2.7 and c) imply that lim My (fn (x, y), K) = My ( lim fn (x, y), K).

n→∞

n→∞

b) implies that the functions fn (x) := My (fn (x, y), K) are continuous, and a) implies that they decrease monotonically to f (x).  2.1.3 We will consider a family of upper semicontinuous functions:{ft : t ∈ T ⊂ (0, ∞)}. It is easy to prove Proposition 2.1.3.1 ft ∈ C + =⇒ inf ft (x) ∈ C + . t∈T

Exercise 2.1.3.1 Prove this proposition. Set fT (x) := sup ft (x). The function fT is not, generally speaking, upper t∈T

semicontinuous even if T is countable and ft are continuous. It is not possible to replace sup in the definition of fT by sup , where T0 is a countable set. However, t∈T

t∈T0

the following theorem holds: Theorem 2.1.3.2 (Choquet’s Lemma) There exists a countable set T0 ⊂ T such that (sup ft )∗ (x) = (supft )∗ (x). t∈T0

t∈T

2.2. Measures and integrals

7

Proof. Let {xn } be a countable set that is dense in Rm and εj ↓ 0. Then the balls Kn,j := {x : |x − xn | < εj } cover every point x ∈ Rm infinitely many times. Renumbering we obtain a sequence {Kl : l ∈ N}. For any l there exists, by definition of sup, such a point x0 ∈ Kl that Kl

supfT (x) ≤ fT (x0 ) + 1/2l.

(2.1.3.1)

Kl

By definition of sup there exists tl such that T

fT (x0 ) < ftl (x0 ) + 1/2l. Thus fT (x0 ) < sup{ftl (x) : x ∈ Kl } + 1/2l.

(2.1.3.2)

The inequalities (2.1.3.1) and (2.1.3.2) imply that for any l there exists tl such that (2.1.3.3) sup{fT (x) : x ∈ Kl } ≤ sup{ftl (x) : x ∈ Kl } + 1/l. Now set T0 = {tl }. Evidently, fT0 (x) ≤ fT (x) and thus fT∗0 (x) ≤ fT∗ (x).

(2.1.3.4)

Let us prove the opposite inequality. Let x ∈ Rm . Choose a subsequence {Klj } that tends to x. From (2.1.3.3) we obtain fT∗ (x) ≤ lim sup sup fT (x ) j→∞

x ∈Klj

≤ lim sup sup ftlj (x ) j→∞

x ∈Klj

(2.1.3.5)

≤ lim sup sup fT0 (x ) = fT∗0 (x). j→∞

x ∈Klj

(2.1.3.4) and (2.1.3.5) imply the assertion of the theorem.



2.2 Measures and integrals 2.2.1 Let G be an open set in Rm and σ(G) a σ-algebra of Borel sets containing all the compact sets K ⊂ G. Let μ be a countably additive nonnegative function on σ(G), which is finite on all K ⊂ G. We will call it a measure or a mass distribution. Let G0 (μ) be the largest open set for which μ is zero. (It is the union of all the open sets G such that μ(G ) = 0.) The set supp μ := G\G0 (μ) is called the support of μ. It is closed in G. We say that μ is concentrated on E ∈ σ(G) if μ(G\E) = 0.

8

Chapter 2. Auxiliary Information. Subharmonic Functions

Theorem 2.2.1.1 (Support) The support of a measure μ is the smallest closed set on which the measure μ is concentrated. Exercise 2.2.1.1 Prove this. A measure μ can be concentrated on a non-closed set E and then E  supp μ. Example 2.2.1.1 Let E be a countable set dense in G. Then supp μ = G and, of course, E = G. The set of all measures on G will be denoted by M(G). The measure μF (E) := μ(E ∩ F ) is called the restriction of μ onto F ∈ σ(G). It is easy to see that μF is concentrated on F and supp μ ⊂ F . A countably additive function ν on σ(G) that is finite for all K ⊂ G is called a charge. We consider only real charges. Example 2.2.1.2 ν := μ1 − μ2 , μ1 , μ2 ∈ M(G). The set of all charges will be denoted Md . Theorem 2.2.1.2 (Jordan decomposition) Let ν ∈ Md (G). Then there exist two sets G+ , G− such that a) G = G+ ∪ G− , G+ ∩ G− = ∅; b) ν(E) ≥ 0 f or E ⊂ G+ ; ν(E) ≤ 0 f or E ⊂ G− . One can find the proof in [Ha, Ch. VI Sec. 29] The measures ν + := νG+ and ν − := νG− , where νG+ , νG− are restrictions of ν to G+ , G− , are called the positive and negative, respectively, variations of ν. The measure |ν| := ν+ + ν− is called the full variation of ν or just a variation. Theorem 2.2.1.3 (Variations) The following holds: ν(E  ); ν = ν + + ν − . ν + (E) = sup ν(E  ); ν − (E) = inf  E ⊂E

E  ⊂E

The proof is easy enough. Exercise 2.2.1.2 Prove this. Example 2.2.1.3 Let ψ(x) be a locally summable function with respect to the Lebesgue measure. Set ν(E) := E ψ(x)dx. Then    + + − − ψ (x)dx, ν (E) = ψ (x)dx; |ν|(E) = |ψ|(x)dx, ν (E) = E

E

E

where ψ + (x) = max(0, ψ(x)); ψ − (x) = − min(0, ψ(x)).

(2.2.1.1)

2.2. Measures and integrals

9

2.2.2 The function f (x), x ∈ G is called a Borel function if the set E A := {f (x) > A} belongs to σ(G) for any A ∈ R. Let K  G be a compact  set and f a Borel function. Then the LebesgueStieltjes integrals of the form K f+ dμ, K f −  dμ with respect to a measure μ ∈ M(G) are defined, and K f dμ := K f + dμ − K f − dμ is defined if at least one of the terms is finite. We say that a property holds μ-almost everywhere on E if the set E0 of x for which it does not hold satisfies the condition μ(E0 ) = 0. We will denote all the compact sets in G as K (sometimes with indexes). The following theorems hold: Theorem 2.2.2.1 (Lebesgue) Let {fn , n ∈ N} be a sequence of Borel functions on K and g(x) ≥ 0 a function on K that is summable with respect to μ (i.e., its integral is finite), |fn (x)| ≤ g(x) for x in K, and fn → f when n → ∞. Then lim K fn dμ = K f dμ. n→∞

Theorem 2.2.2.2 (B. Levy) Let fn ↓ f when n → ∞, and f be a summable function on K.   Then lim K fn dμ = K f dμ. n→∞

Theorem 2.2.2.3 (Fatou’s Lemma) Let fn (x) ≤ const < ∞ for x in K.   Then lim sup K fn dμ ≤ K lim sup fn dμ. n→∞

n→∞

The proofs can be found in [Ha, Ch. V, Sec. 27]. Let L(μ) be the space of functions that are summable with respect to μ. We say that fn → f in L(μ) if fn , f ∈ L(μ) and 

fn − f := |fn − f |(x)dμ → 0 Theorem 2.2.2.4 (Uniqueness in L(μ)) Let fn → f in L(μ) and   fn ψdμ → gψdμ for any ψ continuous on supp μ. Then g − f = 0. For the proof see, e.g., [H¨o, Th. 1.2.5]. 2.2.3 Let φ(x) be a Borel function on G. The set supp φ := {x : φ(x) = 0} is called the support of φ(x). A function φ is called finite in G if supp φ  G. We say  that a sequence μn ∈ M converges weakly to μ ∈ M if the condition  φdμn → φdμ holds for any continuous function φ. We will not show the integration domain, because it is always supp φ. ∗ The weak (it is called also C ∗ -) convergence will be denoted as →.

10

Chapter 2. Auxiliary Information. Subharmonic Functions ∗

Theorem 2.2.3.1 (C ∗ -limits) If μn → μ, then for E ∈ σ(G) the following assertions hold: lim sup μn (E) ≤ μ(E); n→∞

◦

◦

lim inf μn (E) ≥ μ(E); n→∞

◦

where E is the interior of E, E is the closure of E. Proof. Let χE be the characteristic function of the set E. It is upper semicontinuous. Thus there exists a decreasing sequence ϕm of continuous functions finite in G that converges to χE as m → ∞. Then we have   μn (E) = χE dμn ≤ ϕm dμn . Passing to the limit as n → ∞ we obtain  lim sup μn (E) ≤

ϕm dμ.

n→∞

Passing to the limit as m → ∞ we obtain by Theorem 2.2.2.2  lim sup μn (E) ≤ χE dμ = μ(E). n→∞

◦



The proof for E is analogous.

Theorem 2.2.3.2 (Helly) Let {μα : α ∈ A} be a family of measures uniformly bounded on any compact set K ⊂ G, i.e., ∃C = C(K) : μα (K) ≤ C(K), for K  G. Then this family is weakly compact, i.e., there exists a sequence ∗ {αj : αj ∈ A} and a measure μ such that μαj → μ. The proof can be found in [Ha]. A set E is called squarable with respect to measure μ (μ-squarable) if μ(∂E) = 0. Theorem 2.2.3.3. (Squarable Ring) The following holds: sqr1)

if E1 , E2 are μ-squarable, the sets E1 ∩E2 , E1 ∪E2 , E1 \E2 are μ-squarable;

sqr2)

for any couple: an open set G and a compact set K ⊂ G there exists a μ-squarable set E such that K ⊂ E ⊂ G.

Proof. The assertion sqr1) follows from ∂(E1 ∪ E2 )



∂(E1 ∩ E2 )



∂(E1 \E2 ) ⊂ ∂E1 ∪ ∂E2 .

2.2. Measures and integrals

11

Let us prove sqr2). Let Kt := {x : ∃y ∈ K : |x − y| < t} be a t-neighborhood of the K. It is clear that for all the small t we have K  Kt  G. The function a(t) := μ(Kt ) is monotonic on t and thus has no more than a countable set of jumps. Let t be a point of continuity of a(t). Then μ(∂Kt ) ≤ lim [μ(Kt+ ) − μ(Kt− )] = 0. →0



Thus it is possible to set E := Kt for this t. A family Φ of sets is called a dense ring if the following conditions hold: dr1) ∀F1 , F2 ∈ Φ =⇒ F1 ∪ F2 , F1 ∩ F2 ∈ Φ; dr2) ∀K, G : K  G ∃F ∈ Φ : K ⊂ F ⊂ G.

The previous theorem shows that the class of μ-squarable sets is a dense ring. The following theorem shows how one can extend a measure from a dense ring to the Borel algebra. Let Φ be a dense ring and Δ(F ), F ∈ Φ a function of a set which satisfies the conditions: Δ1) monotonicity on Φ: F1 ⊂ F2 =⇒ Δ(F1 ) ≤ Δ(F2 ); Δ2) additivity on Φ: Δ(F1 ∪F2 ) ≤ Δ(F1 )+Δ(F2 ) and Δ(F1 ∪F2 ) = Δ(F1 )+Δ(F2 ) if F1 ∩ F2 = ∅ Δ3) continuity on Φ: ∀F ∈ Φ and > 0 there exists a compact set K and an open set G ⊃ K such that ∀F  ∈ Φ : K ⊂ F  ⊂ G the inequality |Δ(F ) − Δ(F  )| < holds. Theorem 2.2.3.4 (N. Bourbaki) There exists a measure μ such that μ(F ) = Δ(F ), ∀F ∈ Φ iff the conditions Δ1)–Δ3) hold. Theorem 2.2.3.5 (Uniqueness of Measure) Under the conditions Δ1)–Δ3) the measure is defined uniquely by the formulae: μ(K) = inf{Δ(F ) : F ∈ Φ, F ⊃ K}; μ(G) = sup{Δ(F ) : F ∈ Φ, F ⊂ G};

(2.2.3.1) (2.2.3.2)

μ(E) = sup{μ(K) : K ⊂ E} = inf{μ(G) : G ⊃ E},

(2.2.3.3)

and every F ∈ Φ is μ-squarable. For the proof see [Bo, Ch. 4, Sec 3, it. 10]. The squarability follows from (2.2.3.3). The following theorem connects the convergence of measures on any dense ring and on the ring of sets squarable with respect to the limit measure.

12

Chapter 2. Auxiliary Information. Subharmonic Functions

Theorem 2.2.3.6 (Set-convergences) If μn (F ) → μ(F ) for all F in a dense ring Φ, then μn (E) → μ(E) for any μ-squarable set E. ◦

 ∅. Proof. Suppose E = Let > 0. By (2.2.3.3) one can find a compact set K such that ◦

μ(K) + ≥ μ(E) = μ(E).

(2.2.3.4)

One can also find an open set G such that μ(G) − ≤ μ(E) = μ(E).

(2.2.3.5)



By property dr2) of a dense ring one can find F, F ∈ Φ such that ◦

K ⊂ F ⊂ E ⊂ E ⊂ E ⊂ F  ⊂ G. Thus μn (F ) ≤ μn (E) ≤ μn (F  ) and hence μ(F ) ≤ lim μn (E) ≤ lim μn (E) ≤ μ(F  ). n→∞

n→∞

(2.2.3.6)

From (2.2.3.4) and (2.2.3.5) we obtain 0 ≤ μ(F  ) − μ(F ) ≤ μ(G) − μ(K) ≤ 2 for arbitrarily small . Thus from (2.2.3.6) we obtain lim μn (E) = lim μn (E) = μ(E). n→∞

n→∞

(2.2.3.7)

That is to say that μn (E) → μ(E). ◦

If E = ∅, then μ(E) = 0 by the definition of a squarable set. One can show in the same way that μn (E) → 0.  Now we connect the weak convergence to the convergence on squarable sets. Theorem 2.2.3.7 (Set- and C*-convergences) The conditions ∗

μn → μ

(2.2.3.8)

and μn (E) → μ(E) on μ-squarable sets E are equivalent. Proof. Sufficiency of (2.2.3.8) follows from Theorem 2.2.3.1. Exercise 2.2.3.1 Prove this. Let us prove necessity. For any compact set one can find a μ-squarable E such that K ⊂ E. Hence μn (K) ≤ μ(E) + 1 := C(K) when n is big enough. By Helly’s theorem (Theorem 2.2.3.2) there exists a measure μ and a sub∗ sequence μnj → μ . By the proved sufficiency, μ (E) = μ(E) on a dense ring of the squarable sets. Thus μ = μ by Uniqueness Theorem 2.2.3.5. And thus ∗  μn → μ.

2.2. Measures and integrals

Denote by

13

 μE (G) :=

μ(G ∩ E) 0

if G ∩ E = ∅, if G ∩ E = ∅

the restriction of μ on the set E. ∗

∗

Corollary 2.2.3.8 Let μn → μ and E be a squarable set for μ. Then (μn )E → (μ)E . Indeed, if E is a squarable set for μ it is a squarable set for μE . So Theorem 2.2.3.7 implies the corollary. 2.2.4 Let σ(Rm1 × Rm2 ) be the σ-algebra of all the Borel sets, Φi ⊂ σ(Rmi ), i = 1, 2, be dense rings, Φ := Φ1 ⊗ Φ2 ⊂ σ(Rm1 × Rm2 ) be a ring generated by all the sets of form F1 × F2 , Fi ∈ Φi . Theorem 2.2.4.1 (Product of Rings) If Φi , i = 1, 2 are dense rings, then Φ1 ⊗ Φ2 is a dense ring; if they consist of squarable sets, then Φ consists of squarable sets. Proof. Let K ⊂ G ⊂ Rm1 × Rm2 . For every point x ∈ K one can (evidently) find F1 × F2 such that x ⊂ F1 × F2 ⊂ G. One can find a finite covering and obtain a finite union F of sets of such form. Thus F ∈ Φ1 ⊗ Φ2 and F ⊂ G. The second assertion follows from the formula ∂(F1 × F2 ) = (∂F1 × F2 ) ∪ (F1 × ∂F2 ).



Let μi be a measure on σ(Rmi ), i = 1, 2, and μ := μ1 ⊗ μ2 the product of measures, i.e., a measure on σ(Rm1 × Rm2 ) such that μ(E1 × E2 ) = μ1 (E1 )μ2 (E2 ) for all Ei ∈ σ(Rmi ), i = 1, 2. Theorem 2.2.4.2 (Product of Measures) A measure μ1 ⊗ μ2 is uniquely defined by its values on Φ1 ⊗ Φ2 . The assertion follows from Theorem 2.2.4.1 and Uniqueness Theorem 2.2.3.5. Theorem 2.2.4.3 (Fubini) Let f (x1 , x2 ) be a Borel function on Rm1 × Rm2 . Then    f (x1 , x2 )(μ1 ⊗ μ2 )(dx1 dx2 ) = μ1 (dx1 ) f (x1 , x2 )μ2 (dx2 ) Rm1 ×Rm2

R m1

R m2





μ2 (dx2 )

= R m2

f (x1 , x2 )μ1 (dx1 ),

R m1

(2.2.4.1) if at least one of the parts of (2.2.4.1) is well defined. The proof can be found in [Ha, Ch. VII, Sec. 36].

14

Chapter 2. Auxiliary Information. Subharmonic Functions

2.3 Distributions 2.3.1 Let us consider the set D(G) of all infinitely differentiable functions ϕ(x), x ∈ G ⊂ Rm . It is a linear space because for any constants c1 , c2 , ϕ1 , ϕ2 ∈ D(G) =⇒ c1 ϕ1 + c2 ϕ2 ∈ D(G). It is a topological space with convergence defined by ⎧ ⎪ a) supp ϕn ⊂ K  Rm ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ for some compact K D ϕn → ϕ : and ⎪ ⎪ ⎪ b) ϕn → ϕ uniformly on K ⎪ ⎪ ⎪ ⎩ with all their derivatives. We consider some examples of functions ϕ ∈ D. Set  − 1 Ce 1−t2 , for t ∈ (−1; 1), α(t) = 0, for t ∈ (−1; 1).

(D1)

(D2)

(2.3.1.1)

Evidently α(|x|) ∈ D(Rm ) and supp α ⊂ {x : |x| ≤ 1}. Exercise 2.3.1.1 Check this. Let us find C such that   α(|x|)dx = σm

1

α(t)tm−1 dt = 1

(2.3.1.2)

0

where σm is the area of the unit sphere {|x| = 1}. Set  |x| αε (x) := ε−m α . ε

(2.3.1.3)

For any ε we have αε ∈ D and supp αε ⊂ {x : |x| ≤ ε}. Let ψ(y), y ∈ K ⊂ G be a Lebesgue summable function. Consider the function  (2.3.1.4) ψε (x) := ψ(y)αε (x − y)dy. K

The function belongs to D(G) for ε small enough and its support is contained in the ε-neighborhood of K. 2.3.2 Let f (x), x ∈ G ⊂ Rm be a locally summable function in G. The formula  f, ϕ := f (y)ϕ(y)dy, ϕ ∈ D(G) (2.3.2.1)

2.3. Distributions

15

defines a linear continuous functional on D, i.e., one that satisfies the conditions f, c1 ϕ1 + c2 ϕ2  = c1 f, ϕ1  + c2 f, ϕ2 ; D

(ϕn → ϕ) =⇒ f, ϕn  → f, ϕ.

(D 1) (D 2)

However, (2.3.2.1) does not exhaust all the linear continuous functionals as we will see further. An arbitrarily linear continuous functional on D(G) is called a Schwartz distribution and the linear topological space of the functionals is denoted as D (G). Following are some examples of functionals that do not have the form of (2.3.2.1): (2.3.2.2) δx , ϕ := ϕ(x). This distribution is called the Dirac delta-function. Further, δx(n) , ϕ := (−1)n ϕ(n) (x).

(2.3.2.3)

This distribution is called the nth derivative of the Dirac delta-function. Exercise 2.3.2.1 Check that the functionals (2.3.2.2) and (2.3.2.3) are both distributions. A distribution of the form (2.3.2.1) is called regular. Theorem 2.3.2.1 (Du Bois Reymond) If two locally summable functions f1 and f2 define the same distribution, then they coincide almost everywhere. For the proof see, e.g., [H¨o, Thm. 2.1.6]. Note that the converse assertion is obvious. A distribution μ is called positive if μ, ϕ ≥ 0 for any ϕ ∈ D(G) such that ϕ(x) ≥ 0 for all x ∈ Rm . We shall write this as μ > 0 in D . Example 2.3.2.1 Let μ(E) be a measure in G. Then the distribution  μ, ϕ := ϕ(x)μ(dx)

(2.3.2.4)

is positive. This formula represents all the positive distributions as one can see from Theorem 2.3.2.2 (Positive Distributions) Let μ > 0 in D(G). Then there exists a unique measure μ(E) such that the distribution μ is given by (2.3.2.4). For the proof see, e.g., [H¨o, Thm. 2.1.7]. 2.3.3 Let us consider operations on distributions. A product of a distribution f by an infinitely differentiable function α(x) is defined by αf, ϕ := f, αϕ. (2.3.3.1) It is well defined because αϕ ∈ D too.

16

Chapter 2. Auxiliary Information. Subharmonic Functions

A sum of distributions f1 and f2 is defined by f1 + f2 , ϕ := f1 , ϕ + f2 , ϕ, and the partial derivative

∂ ∂xk



(2.3.3.2)

is defined by the equality

∂ ∂ f, ϕ := f, − ϕ. ∂xk ∂xk

(2.3.3.3)

These definitions look reasonable because of the following Theorem 2.3.3.1 (Operations on Distributions) The sum of regular distributions corresponds to the sum of the functions; the product of a regular distribution by an infinitely differentiable function corresponds to the product of the functions; the derivative of a regular distribution that is generated by a differentiable function corresponds to the derivative of that function. Proof. We have, for example,   α · (f ), ϕ := f (x)[α(x)ϕ(x)]dx = [α(x)f (x)]ϕ(x)dx := (αf ), ϕ. For the sum we have



(f1 ) + (f2 ), ϕ := f1 , ϕ + f2 , ϕ =



∂ ∂xk f .

f2 (x)ϕ(x)dx



= Let f (x) have the derivative

 f1 (x)ϕ(x)dx +

[f1 (x) + f2 (x)]ϕ(x)dx = (f1 + f2 ), ϕ.

Then

∂ ∂ f, ϕ : = f, − ϕ ∂xk ∂xk

  ∂ = f (x1 , x2 , . . . , xm ) − ϕ(x1 , x2 , . . . , xm ) dx1 dx2 , . . . , dxm ∂xk  = dx1 , . . . , dxk−1 dxk+1 , . . . , dxm

  ∂ × f (x1 , x2 , . . . , xm ) − ϕ(x1 , x2 , . . . , xm ) dxk . ∂xk

Now we shall integrate by parts and all the substitution will vanish, because ϕ is finite. So we obtain  ∂ ∂  f, ϕ = f (x)ϕ(x)dx. ∂xk ∂xk That is to say the derivative of the distribution corresponds to the function derivative. 

2.3. Distributions

17

2.3.4 We say that a sequence of distributions fn converges to a distribution f if fn , ϕ → f, ϕ ∀ϕ ∈ D(G).

(2.3.4.1)

Theorem 2.3.4.1 (Completeness of D ) If the sequence of numbers fn , ϕ has a limit for every ϕ ∈ D(G), then this functional is a linear continuous functional on D(G), i.e., a distribution. For the proof see, e.g., [H¨o, Thm. 2.1.8]. Differentiation is continuous with respect to convergence of distributions. Theorem 2.3.4.2 (Continuity of Differential Operators) If fn → f in D(G), then ∂ ∂ ∂xk fn → ∂xk f . Proof. Set in (2.3.4.1) ϕ := − ∂x∂ k ϕ. Then         ∂ ∂ ∂ ∂ fn , ϕ = fn , − ϕ → f, − ϕ = f, ϕ . ∂xk ∂xk ∂xk ∂xk



The following theorem shows that the D -convergence is the weakest of the convergences considered earlier. Theorem 2.3.4.3 (Connection between Convergences) Let fn be a sequence of Lebesgue summable functions on domain G such that at least one of the following conditions holds: Cnvr1) fn → f uniformly on any compact set K  G and f is a locally summable function; Cnvr2) fn → f on any K  G, satisfying the conditions of the Lebesgue theorem (Theorem 2.2.2.1); Cnvr3) fn ↓ f monotonically and f is a locally summable function. Then fn → f in D (G). Proof. All the assertions are corollaries of Section 2.2.2 on passing to the limit under an integral. Let us prove, for example, Cnvr3). Let fn ↓ f . Then    (2.3.4.2) fn , ϕ = fn (x)ϕ(x)dx = fn (x)ϕ+ (x)dx − fn (x)ϕ− (x)dx where ϕ+ and ϕ− are defined in (2.2.1.1). Both last integrals in (2.3.4.2) have a limit by the B. Levy theorem (Theorem 2.2.2.2), and thus    lim fn , ϕ = f (x)ϕ+ (x)dx − f (x)ϕ− (x)dx = f (x)ϕ(x)dx = f, ϕ. n→∞

(2.3.4.3) means that fn → f in D .

(2.3.4.3) 

18

Chapter 2. Auxiliary Information. Subharmonic Functions

Exercise 2.3.4.1 Prove Cnvr 1) and 2). Theorem 2.3.4.4 (D and C ∗ convergences) Let μn , μ be measures in G. The con∗ ditions μn → μ in D (G) and μn → μ are equivalent. It is clear that the first condition is necessary for the second one. The sufficiency holds, because every continuous function can be approximated with functions that belong to D. For more details see, e.g., [H¨ o, Thm. 2.1.9]. Let α (x) be defined as in (2.3.1.3).For any f ∈ D (D) we can consider the function f (x) := f, α (x + •). It is called a regularization of the distribution f . Theorem 2.3.4.5.(Properties of Regularizations) The following holds: reg1)

f (x) is an infinitely differentiable function in any K  D for sufficiently small ;

reg2)

f (x) → f in D (D) as ↓ 0;

reg3)

if fn → f in D (D), (fn ) → f uniformly with all its derivatives on any compact set in D. The property reg1) follows from the formula   ∂ ∂ f = f, α (x + •) . ∂xj ∂xj

The property reg2) follows from the assertion  φ (x) := φ(y)α (x + y)dy → φ(x) in D(D) as ↓ 0. For the proof of reg3) see [H¨ o, Theorems 2.1.8, 4.1.5]. Let us note the following assertion; Theorem 2.3.4.6 (Continuity •, •) The function f, φ : D (G) × D(G) → R is continuous in the appropriate topology. I.e., fn → f in D (G) and φj → φ in D(G) imply fn , φj  → f, φ. For the proof see [H¨ o, Theorem 2.1.8]. 2.3.5 Let G1 ⊂ G. Then D (G) ⊂ D (G1 ), because every functional on D(G) can be considered as a functional on D(G1 ). A distribution f ∈ D (G) considered as a distribution in D (G1 ) is called the restriction of f to G1 and is denoted f |G1 .

2.3. Distributions

19

Theorem 2.3.5.1 (Sewing Theorem) Let Gα ⊂ Rm be a family of domains and in every of them let there be a distribution fα ∈ D(Gα ), such that: If Gα1 ∩ Gα2 = ∅, the equality fα1 |Gα1 ∩Gα2 = fα2 |Gα1 ∩Gα2

(2.3.5.1)  holds. Then there exists one and only one distribution f ∈ D (G) where G = Gα α

such that f |Gα = fα .

In particular, it means that every distribution is defined uniquely by its restriction to a neighborhood of every point. Let D(SR ) be a space of infinitely differentiable functions on the sphere SR := {x : |x| = R}. The corresponding distribution space is denoted as D (SR ). The sewing theorem holds for this space in the following form: Theorem 2.3.5.2 (D on Sphere) Let a family of domains Ωα cover SR and in every of them let there be a distribution fα ∈ D(Ωα ), such that: If Ωα1 ∩ Ωα2 = ∅, the equality fα1 |Ωα1 ∩Ωα2 = fα2 |Ωα1 ∩Ωα2

(2.3.5.2)

holds. Then there exists one and only one distribution f ∈ D (SR ) such that f |Ωα = fα . 2.3.6 Let L :=

 ∂ ∂ ai,j (x) + q(x) ∂x ∂x i j i,j

(2.3.6.1)

be a differential operator of second order with infinitely differentiable coefficients ai,j , q. We will consider only three types of differential operators: a one-dimensional operator with constant coefficients,the Laplace operator and the so-called spherical operator (see Section 2.4). For all these operators we have the following assertion which follows from the general theory (see, e.g., [H¨ o, Theorem 11.1.1]): Theorem 2.3.6.1 (Regularity of Generalized Solution) If the equation Lu = 0 has a solution u ∈ D (G), then u is a regular distribution and can be realized as an infinitely differentiable function. A distribution that satisfies the equation Lu = δy in D (G),

(2.3.6.2)

where δy is a Dirac delta function (see (2.3.2.2)), is called a fundamental solution of L at the point y.

20

Chapter 2. Auxiliary Information. Subharmonic Functions

Every differential operator that we are going to consider has a fundamental solution (see, e.g., [H¨ o, Theorem 10.2.1]). A restriction of the equation (2.3.6.2) to the domain Gy := G\y is a homogeneous equation Lu = 0 in D (Gy ). Thus we have Theorem 2.3.6.2 (Regularity of Fundamental Solution) The fundamental solution is an infinitely differentiable function outside the point y. 2.3.7 We will need further also the Fourier coefficients for the distribution on the circle. Let D(S 1 ) be a set of all infinitely differentiable functions on the unit circle 1 S . The set of all linear continuous functionals over D(S 1 ) with the corresponding topology (see 2.3.2) is the corresponding space of distributions D (S 1 ) for which all the previous properties of distributions holds. The functions {eikφ , k = 0, ±1, ±2, . . . } belong to D(S 1 ). The Fourier coefficients of ν ∈ D (S 1 ) are defined by νˆ(k) := ν, e−ikφ .

(2.3.7.1)

The inverse operator is defined by ν, g =

∞ 1  νˆ(k)g, eikφ , 2π

(2.3.7.2)

k=−∞

and the series converges, in any case, for those ν that are finite derivatives of summable functions, because Fourier coefficients of g decrease faster then every power of x. The convolution of distribution ν ∈ D (S 1 ) and g ∈ D(S 1 ) is defined by ν ∗ g(φ) = ν, g(φ − •).

(2.3.7.3.)

This is a function from D(S 1 ). The convolution of distributions ν1 , ν2 ∈ D (S 1 ) is defined by ν1 ∗ ν2 , g = ν1 ∗ (ν2 ∗ g).

(2.3.7.4)

In spite of the view it is commutative and ˆ1 (k) · νˆ2 (k). ν 1 ∗ ν2 (k) = ν Exercise 2.3.7.1 Count the Fourier coefficients of the functions G(reiφ ) = log |1 − reiφ |

(2.3.7.5)

for r > 1, r = 1, r < 1; the function defined by c os ρ(φ) := cos ρφ, −π < φ < π, ρ ∈ (0, ∞)

(2.3.7.6)

2.4. Harmonic functions

21

and 2π-periodically extended; the function φ˜ sin pφ, p ∈ N

(2.3.7.7)

where φ˜ is the 2π-periodical extension of the function f (φ) = φ, φ ∈ [0, 2π). Exercise 2.3.7.2 Set Pp−1 (re ) :=  iφ

 p−1  rk eikφ k=1

k

, p ∈ N.

(2.3.7.8)

Prove that for every distribution ν : i• ) ∗ ν)(p) = 0. (Pp−1 (re

(2.3.7.9)

The same for the function Gp (reiφ ) := G(reiφ ) + Pp (reiφ ) for r < 1.

2.4 Harmonic functions 2.4.1 We will denote as Δ the Laplace operator in Rm : Δ :=

∂2 ∂2 + · · · + . ∂x21 ∂x2m

We introduce in Rm the spherical coordinate system by the formulae: x1 = r sin φ0 sin φ1 . . . sin φm−2 ; x2 = r cos φ0 sin φ1 . . . sin φm−2 ; x3 = r cos φ1 sin φ2 . . . sin φm−2 ; ............ xk = r cos φk−2 sin φk−1 . . . sin φm−2 ; ............ xm = r cos φm−2 , where 0 < φ0 ≤ 2π; 0 ≤ φj < π, j = 1, m − 2; 0 < r < ∞. Passing to the coordinates (r, φ0 , φ1 , . . . , φm−2 ) in the Laplace operator we obtain 1 ∂ 1 ∂ + 2 Δx0 . Δ = m−1 rm−1 r ∂r ∂r r

22

Chapter 2. Auxiliary Information. Subharmonic Functions

The operator Δx0 is called spherical, and has the form Δx0 :=

m−2  i=0

where Π :=

m−2 

sinj φj ; Πi :=

j=1

1 ∂ Π ∂ , Π ∂φi Πi ∂φi m−2 

sin2 φj ; Πm−2 := 1.

j=i+1

In particular, for m = 2, i.e., for the polar coordinates, Δ=

1 ∂2 1 ∂ ∂ r + 2 2. r ∂r ∂r r ∂φ

A distribution H ∈ D (G) is called harmonic if it satisfies the equation ΔH = 0. The next theorem follows from Theorem 2.3.6.1. Theorem 2.4.1.1 (Smoothness of harmonic functions) Any harmonic distribution is equivalent to an infinitely differentiable function. This function, of course, satisfies the same equation and is a harmonic function in the ordinary sense. A direct proof can be found, e.g., in [Ro, Ch. 1, § 2 (1.2.5), p. 60]. Let f (z), z = x + ıy be a holomorphic function in a domain G ⊂ C. Then the functions u(x, y) := f (z) and v(x, y) := f (z) are harmonic in G. In particular, the functions rn cos nϕ and rn sin nϕ where r = |z|, ϕ = arg z are harmonic. Set  −|x|2−m , for m ≥ 3, Em (x) := (2.4.1.1) log |z|, for m = 2. (We will often denote points of the plane as z.) It is easy to check that Em (x) is a harmonic function for |x| = 0. Set  for m ≥ 3; (m − 2)σm , θm := 2π, for m = 2, where σm is the surface area of the unit sphere in Rm . Theorem 2.4.1.2 (Fundamental Solution) The function Em (x − y) satisfies in D (Rm ) the equation1 Δx Em (x − y) = θm δ(x − y), (2.4.1.2) where δ(x) is the Dirac δ-function (see 2.3.2). 1E m

is slightly different from the fundamental solution (see, (2.3.6.2)), but this is traditional in Potential Theory

2.4. Harmonic functions

23

Proof. Let us prove the equality (2.4.1.2) for y = 0. Suppose φ ∈ D(Rm ) and supp φ ⊂ K  Rm . We have   Em (x)Δφ(x)dx. ΔEm , φ := Em (x)Δφ(x)dx = lim →0 |x|≥

Transforming this integral by the Green formula and using the fact that φ is finite we obtain     ∂φ ∂Em ds, Em (x)Δφ(x)dx = ΔEm (x)φ(x)dx + Em ds − φ ∂n ∂n |x|≥

|x|≥

|x|=

|x|=

∂ is the differentiation in the direction where ds is an element of surface area and ∂n of the external normal. Use the harmonicity of Em . Then the first integral is equal to zero. Further we have      ∂φ ∂ 0 φ(rx )ds  Em ds = = O( ), for → 0. ∂n ∂r |x|=

For the third term we have   m − 2 m−1 ∂Em ds = m−1 φ ∂n

|x|=

r=

|x0 |=1

φ(rx0 )ds = [φ(0) + o(1)](m − 2)σm .

|x0 |=1

Thus we obtain ΔEm , φ = φ(0)θm , and this proves (2.4.1.2) for y = 0. It is clear that by changing φ(x) for φ(x+y) we obtain (2.4.1.2) in the general case.  We will consider now a domain Ω with a Lipschitz boundary (Lipschitz domain). It means that every part of ∂Ω can be represented in some local coordinates (x, x ), x ∈ R, x ∈ Rm−1 in the form x = f (x ), where f is a Lipschitz function, i.e., |f (x1 ) − f (x2 )| ≤ M∂Ω |x1 − x2 | where M depends only on the whole ∂Ω and does not depend on this local part. Let G(x, y, Ω) be the Green function of a Lipschitz domain Ω. It is known (see, e.g., [Vl, Ch. V, §28]) that the Green function has the following properties: G(x, y, Ω) < 0, for (x, y) ∈ Ω × Ω; G(x, y, Ω) = 0 for (x, y) ∈ Ω × ∂Ω;

(g1)

G(x, y, •) = G(y, x, •);

(g2)

G(x, y, •) − Em (x − y) = H(x, y),

(g3)

where H is harmonic on x and on y within Ω; −G(x, y, Ω1 ) ≤ −G(x, y, Ω2 ) for Ω1 ⊂ Ω2 .

(g4)

24

Chapter 2. Auxiliary Information. Subharmonic Functions

From (g3) follows Theorem 2.4.1.3 (Green Function) The equality Δx G(x, y, Ω) = θm δ(x − y),

(2.4.1.3)

holds in D (Ω). Let f (x) be a continuous function on ∂Ω. It is known (see, e.g., [Vl, Ch. V, § 29]) that the function  ∂ (2.4.1.4) H(x, f ) := f (y) G(x, y, Ω)dsy ∂n ∂Ω

is the only harmonic function that coincides with f on ∂Ω. The unique solution of the Poisson equation Δu = p, u|∂Ω = f for a continuous function p is given by the formula   ∂ −1 u(x, f, p) := f (y) G(x, y, Ω)dsy + θm G(x, y, Ω)p(y)dy. ∂ny

(2.4.1.5)

Ω

∂Ω

Let D be an arbitrarily open domain.We can define a G(x, y, D) in the following way. Consider a sequence Ωn of a Lipschitz domain such that Ωn ↑ D. The sequence of the corresponding Green functions G(x, y, Ωn ) monotonically decreases. If it is bounded from below in some point, it is bounded everywhere while x = y (as it follows from Theorem 2.4.1.7). It can be shown that the limit exists for any domain, the boundary of which has positive capacity (see 2.5 and references there). We will mainly use the Green function for the Lipschitz domains. Let G(x, y, Ka,R ) be the Green function of the ball Ka,R := {|x − a| < R}. Theorem 2.4.1.4 (Green Function for a Ball)  ∗ |y−a||x−ya,R | 2−m ) , −|x − y|2−m − ( R G( x, y, Ka,R ) = |ζ−z|R log |ζ−a||z−ζ ∗ | ∗ ya.R



where := a + (y − a) R {|x − a| = R}.

2



a.R

|y − a|

2



for m ≥ 3, for m = 2,

is the inversion of y relative to the sphere

For the proof see, e.g., [Br, Ch. 6, § 3]. Theorem 2.4.1.5 (Poisson Integral) Let H be a harmonic function in Ka,R and continuous in its closure. Then  R2 − |x − a|2 1 H(y) dsy , x ∈ K(a, R). (2.4.1.6) H(x) = σm R |x − y|m |x−a|=R

2.4. Harmonic functions

25

In particular, for m = 2, 1 H(a + re ) = 2π

2π

ıφ

H(a + Reıψ ) 0

R2

R2 − r 2 dψ. − 2Rr cos(φ − ψ) + r2

This theorem follows from (2.4.1.4). Theorem 2.4.1.6 (Mean Value) Let H be harmonic in G ⊂ Rm . Then  1 H(y)dsy , H(x) = σm Rm−1

(2.4.1.7)

|x−a|=R

where x ∈ G and R is taken such that K(x, R)  G. We must only set a := x in (2.4.1.6). We can rewrite (2.4.1.7) in the form  1 H(x) = H(x + Ry)dsy . σm |y|=1

Theorem 2.4.1.7 (Harnack) Suppose the family {Hα ), α ∈ A} of harmonic functions in G satisfies the conditions Hα (x) ≤ C(K), for x ∈ K; Hα (x0 ) ≥ B > −∞, for x0 ∈ K

(Har1) (Har2)

for every compact K  G and C(K), B are constants not depending on α. Then the family is precompact in the uniform topology, i.e., there exists such a sequence Hαn , and a function H harmonic in the interior of K and continuous in K such that Hαn → H uniformly in every K. One can prove by using (2.4.1.6) that | grad Hα | are bounded on every compact set by a constant not depending on α. Thus the family is uniformly continuous and thus it is precompact by the Ascoli theorem. For details see, e.g., [Br, Supplement, § 7]. Theorem 2.4.1.8 (Uniform and D -convergences) Suppose the sequence Hn satisfies the conditions of the Harnack theorem and converges to a function H in D (G). Then Hn converges to H uniformly on every compact set K  G. Of course, H is harmonic in G. Proof. By the Harnack theorem the family is precompact. Thus we must only prove the uniqueness of H. Suppose there exist two subsequences such that Hk1 → H 1 and Hk2 → H 2 uniformly on every compact K  G. By Connection between Convergences (Theorem 2.3.4.3) Hk1 → H 1 and 2 Hk → H 2 in D . Hence, H 1 = H 2 in D (G). By the De Bois Raimond theorem (Theorem 2.3.2.1) H 1 = H 2 almost everywhere and hence everywhere because these functions are continuous. 

26

Chapter 2. Auxiliary Information. Subharmonic Functions

Let D be a domain with a smooth boundary ∂D and let F ⊂ ∂D. Set  ∂G ω(x, F, D) := (x, y)dsy . ∂ny F

It is called a harmonic measure of F with respect to D. A harmonic measure can be defined for an arbitrary domain D by a limit process similar to the one we had for the Green function. In this case the formula (2.4.1.4) has the form  f (y)dω(x, y, D). H(x, f ) := ∂D

However we can not assert that H(x, f ) coincides with f in any point x ∈ ∂D. We can only consider it as an operator that maps a function defined on ∂D to a harmonic function in D. By (2.4.1.3) we obtain Theorem 2.4.1.9 (Two Constants Theorem) Let H be harmonic in D and satisfy the conditions H(x) ≤ A1 for x ∈ F ; H(x) ≤ A2 for x ∈ ∂D\F where A1 and A2 are constants. Then H(x) ≤ A1 ω(x, F, D) + A2 ω(x, ∂D\F, D) for x ∈ D. ∗ Let ya,R be the inversion from Green Function for a Ball (Theorem 2.4.1.4). ∗ ∗ , i.e., the inversion relative to a unit sphere with the center in the Set y := y0,1 origin. Let G∗ := {y ∗ : y ∈ G} be the inversion of a domain G.

Theorem 2.4.1.10 (Kelvin’s Transformation) If H is harmonic in G, then H ∗ (y) := |y|2−m H(y ∗ )

(2.4.1.8)

is harmonic in G∗ . For the proof you must honestly compute Laplacian of H ∗ . “The computation is straightforward but tedious” ([He, Thm. 2.24]). It is not so tedious if you use the spherical coordinate system. Exercise 2.4.1.1 Do this. 2.4.2 Denote as S1 := {x0 : |x0 | = 1} the unit sphere with center in the origin. A function Yρ (x0 ), x0 ∈ Ω ⊂ S1 is called a spherical function of degree ρ if it satisfies the equation (2.4.2.1) Δx0 Y + ρ(ρ + m − 2)Y = 0. For m = 2, (2.4.2.1) gets the form Y  (θ) + ρ2 Y (θ) = 0,

i.e., Y (θ) = a cos ρθ + b sin ρθ.

Spherical functions are obtained if we solve the equation ΔH = 0 by the change H(x) = |x|ρ Y (x0 ).

2.4. Harmonic functions

27

Theorem 2.4.2.1 (Sphericality and Harmonicity) The function Yρ (x0 ) is spherical in a domain Ω ⊂ S1 if and only if the functions H(x) = |x|ρ Yρ (x0 ) and H ∗ (x) = |x|−ρ−m+2 Yρ (x0 ) are harmonic in the cone Con(Ω) := {x = rx0 : x0 ∈ Ω, 0 < r < ∞}.

(2.4.2.2)

If ρ = k, k ≥ 0, k ∈ Z, and only in this case, Yk (x0 ) is spherical on the whole S1 , H(x) is a homogeneous harmonic polynomial of degree k and H ∗ is harmonic in Rm \0. For the proof see, e.g., [Ax, Ch. 5] The spherical functions of an integer degree k form a finite-dimension space of dimension (2k + m − 2)(k + m − 3)! . dim(m, k) = (m − 2)!k! In particular, d(2, k) = 2 for any k. For different k the spherical functions Yk (x0 ) are orthogonal on S1 . In particular, for m = 2, it means the orthogonality of the trigonometric functions system. Theorem 2.4.2.2 (Expansion of a Harmonic Function) Let H(x) be a harmonic function in the ball KR := {|x| < R}. There exists an orthonormal system of spherical functions Yk (x0 ), k = 0, ∞, depending on H such that H(x) =

∞ 

ck Yk (x0 )|x|k , for |x| < R.

(2.4.2.3)

k=0

For any such system we have 1 ck = k R

 H(Rx0 )Yk (x0 )dsx0 .

(2.4.2.4)

S1

For the proof see, e.g., [Ax, Ch. 10], [TT, Ch. 4, § 10]. Theorem 2.4.2.3 (Liouville) Let H be harmonic in Rm and suppose lim inf R−ρ max H(x) < ∞ R→∞

|x|=R

(2.4.2.5)

holds. Then H is a polynomial of a degree q ≤ ρ. Proof. We can suppose H(0) = 0 because H(x) − H(0) is harmonic and also satisfies (2.4.2.5). Let Rn → ∞ be a sequence for which Rn−ρ max H(x) ≤ const < ∞. |x|=Rn

(2.4.2.6)

28

Chapter 2. Auxiliary Information. Subharmonic Functions

From (2.4.2.4) we obtain |ck | ≤ Ak R−k

 |H(Rx0 )|dsx0 ,

(2.4.2.7)

S1

where Ak = maxS1 |Yk (x0 )|. From the mean value property (Theorem 2.4.1.6)  H(Rx0 )dsx0 = H(0)σm = 0. S1

Thus



 |H(Rx0 )|dsx0 = 2

S1

H + (Rx0 )dsx0 ≤ 2σm max H(x). |x|=R

(2.4.2.8)

S1

From (2.4.2.8) and (2.4.2.7) we have |ck | ≤ 2Ak R−k σm max H(x). |x|=R

(2.4.2.9)

Set R := Rn and k > ρ. Passing to the limit when n → ∞, we obtain ck = 0 for k > ρ. Then (2.4.2.3.) implies that H is a harmonic polynomial of degree q ≤ ρ. 

2.5 Potentials and capacities 2.5.1 Let G(x, y.D) be the Green function of a Lipschitz domain D. We will suppose it is extended as zero outside of D.  Π(x, μ, D) := − G(x, y, D)μ(dy) is called the Green potential of μ relative to D. The domain of integration will always be Rm . Theorem 2.5.1.1 (Green Potential Properties) The following holds: GPo1) Π(x, μ, D) is lower semicontinuous; GPo2) it is summable over any (m − 1)-dimensional hyperplane or smooth hypersurface; GPo3) ΔΠ(•, μ, D) = −θm μ in D (D); GPo4) the reciprocity law holds:   Π(x, μ1 , D)μ2 (dx) = Π(x, μ2 , D)μ1 (dx).

2.5. Potentials and capacities

29

GPo5) semicontinuity in μ: if μn → μ in D (Rm ), then lim inf Π(x, μn , D) ≥ Π(x, μ, D). n→∞

GPo6) continuity in μ in D : if μn → μ, then Π(•, μn , D) → Π(•, μ, D) in D (Rm ) and in D (SR ), where SR is the sphere {|x| = R}. Proof. Let us prove GPo1). Let N > 0. Set GN (x, y) := max(G(x, y), −N ), a truncation of the function G(x, y). The functions GN are continuous in Rm × Rm and GN (x, y) ↓ G(x, y) for every (x, y) when N → ∞. Set  ΠN (x, μ, D) := − GN (x, y, D)μ(dy). The functions ΠN are continuous and ΠN (x, •) ↑ Π(x, •) by the B. Levy theorem (Theorem 2.2.2.2). Then ΠN (x, •) is lower semicontinuous by the Second Criterion of semicontinuity (Theorem 2.1.2.9). Let us prove GPo5). From Theorem 2.3.4.4 (D and C ∗ convergences) lim ΠN (x, μn , D) = ΠN (x, μ, D).

n→∞

Further Π(x, μn , D) ≥ ΠN (x, μn , D), hence lim inf Π(x, μn , D) ≥ ΠN (x, μ, D). n→∞

Passing to the limit while N → ∞, we obtain GPo5). The assertion GPo2) follows from the local summability of the function |x|2−m that can be checked directly. Let us prove GPo3). For φ ∈ D(D) we have   ΔΠ, φ := Π, Δφ = − μ(dy) G(x, y, D)Δφ(x)dx   = − Δx G(•, y, D), φμ(dy) = −θm φ(y)μ(dy) = −θm μ, φ, since Δx G(•, y, D), φ = θm φ(y) by Theorem 2.4.1.3. The property GPo4) follows from the symmetry of G(x, y, •) (property (g2)).  Let us prove GPo6). Note that integral |x|m−1 dx converges locally in Rm and in Rm−1 . From this one can obtain by some simple estimates that functions

30

Chapter 2. Auxiliary Information. Subharmonic Functions

  Ψ(y) := G(x, y, D)ψ(x)dx while ψ ∈ D(Rm ) and Θ(y) := SR G(x, y, D)θ(x)dsx while θ ∈ D(SR ) are continuous on y ∈ Rm . Now we have   Π(•, μn , D), ψ = Ψ(y)μn (dy) → Ψ(y)μ(dy) = Π(•, μ, D), ψ. Thus the first assertion in GPo6) is proved. The second one can be proved in the same way.  Set ν := μ1 −μ2 , and let Π(x, ν, D) := Π(x, μ1 , D)−Π(x, μ2 , D) be a potential of this charge. Consider the boundary problem of the form Δu = μ1 − μ2 , in D (D),

u|∂D = f,

(2.5.1.2)

where f is a continuous function. Theorem 2.5.1.2 (Solution of Poisson Equation) The solution of the boundary problem (2.5.1.2) is given by the formula −1 Π(x, ν, D), u(x) = H(x, f ) − θm

where H(x, f ) is the harmonic function from (2.4.1.4). Proof. Since Π(x, ν, D)|∂D = 0, the function u(x) satisfies the boundary condition. Using GPo3) we obtain Δu = ΔH − [θm ]−1 ΔΠ = μ1 − μ2 . A potential of the form



Π(x, μ) :=



μ(dy) |x − y|m−2

is called a Newton potential. It is the Green potential for D = Rm . The potential  Π(z, μ) = − log |z − ζ|μ(dζ) is called logarithmic. 2.5.2 Let K  D. The quantity capG (K, D) := sup μ(K)

(2.5.2.1)

where the supremum is taken over all mass distributions μ for which the following conditions are satisfied: Π(x, μ, D) ≤ 1, supp μ ⊂ K,

(2.5.2.2) (2.5.2.3)

is called the Green capacity of the compact set K relative to the domain D.

2.5. Potentials and capacities

31

Theorem 2.5.2.1 (Properties of capG ) For capG the following properties hold: capG1)

monotonicity with respect to K: K1 ⊂ K2 implies capG (K1 , D) ≤ capG (K2 , D).

capG2)

monotonicity with respect to D: K  D1 ⊂ D2 implies capG (K, D1 ) ≥ capG (K, D2 )

capG3)

subadditivity with respect to K: capG (K1 ∪ K2 , D) ≤ capG (K1 , D) + capG (K2 , D).

Proof. The set of all mass distributions that satisfy (2.5.2.2) for K = K1 is not less than the analogous set for K = K2 . Thus capG1) holds. By the Green function property (g3) (see § 2.4.1) −G(x,y,D1 ) ≤ −G(x,y,D2 ). Thus the set of all μ that satisfy (2.5.2.2) for D = D1 is wider than for D = D2 . Hence capG2) holds. Let supp μ ⊂ K1 ∪ K2 and let μ1 , μ2 be the restrictions of μ to K1 , K2 respectively. If μ satisfies (2.5.2.2) for K := K1 ∪ K2 then μ1 , μ2 satisfy (2.5.2.2) for K := K1 , K2 respectively. From the inequality μ(K1 ∪ K2 ) ≤ μ(K1 ) + μ(K2 ) we obtain that μ(K1 ∪ K2 ) ≤ capG (K1 , D) + capG (K2 , D) for any μ with supp μ ⊂ K1 ∪ K2 . Thus capG3) holds. The equivalent definition of the Green capacity is given by Theorem 2.5.2.2 (Dual Property) The following holds: capG (K, D) = [inf sup Π(x, μ, D)]−1 μ x∈D

(2.5.2.4)

where the infimum is taken over all mass distributions μ such that μ(K) = 1. For the proof see, e.g., [La, Ch. 2, § 4 it. 18]. For D = Rm , m ≥ 3, the Green capacity is called Wiener capacity (capm (K)). It has the following properties in addition to those of the Green capacity: capW1) invariance with respect to translations and rotations, i.e., capm (V (K + x0 )) = capm (K), where V K and K + x0 are the rotation and the translation of K respectively. The presence of these properties brings the notion of capacity closer to the notion of measure. Thus it is natural to extend the capacity to the Borel algebra of sets.

32

Chapter 2. Auxiliary Information. Subharmonic Functions

The Wiener capacity of an open set is defined as capm (D) := sup capm (K), K

where the supremum is taken over all compact K  D. The outer and inner capacity of any set E can be defined by the equalities capm (E) := inf capm (D); capm (E) := sup capm (K). D⊃E

K⊂E

A set E is called capacible if capm (E) = capm (E). Theorem 2.5.2.3 (Choquet) Every set E belonging to the Borel ring is capacible. For the proof see, e.g., [La, Ch2, Thm. 2.8]. Sets which have “small size” are sets of zero capacity. We emphasize the following properties of these sets: j capZ1) If capm (E j ) = 0, j = 1, 2, . . . . then capm (∪∞ 1 E ) = 0;

capZ2) Having the property of zero capacity does not depend on the type of capacity: Green, Wiener or logarithmic capacity that we define below. Example 2.5.2.1 Using Theorem 2.5.2.2 we obtain that any point has zero capacity, because for every mass distribution concentrated in the point the potential is equal to infinity. The same holds for any set of zero m − 2 Hausdorff measure (see 2.5.4). Example 2.5.2.2 Any (m − 1)-hyperplane or smooth hypersurface has positive capacity, because the potential with masses uniformly distributed over the surface is bounded. The Wiener 2-capacity can be defined naturally only for sets with diameter less then 1, because the logarithmic potential is positive only when this condition holds. Instead, one can use the logarithmic capacity which is defined by the formulae capl (K) := exp[−cap2 (K)] for K ⊂ {|z| < 1} and

(2.5.2.5)

capl (K) := t−1 capl (tK)

for any other bounded K, where t is chosen in such a way that tK ⊂ {|z| < 1}. One can check that this definition is correct, i.e., it does not depend on t. 2.5.3 Theorem 2.5.3.1 (Balayage; sweeping) Let D be a domain such that ∂D  Rm , and supp μ  D. Then there exists a mass distribution μb such that for m ≥ 3, or for m = 2 and for D which is a bounded domain, the following holds: bal1)

Π(x, μb ) < Π(x, μ) for x ∈ D;

bal2)

/ D; Π(x, μb ) = Π(x, μ) for x ∈

2.5. Potentials and capacities

bal3)

supp μb ⊂ ∂D;

bal4)

μb (∂D) = μ(D).

33

If m = 2 and the domain is unbounded, a potential of the form  ˆ μ) := − log |1 − z/ζ|μ(dζ) Π(z, satisfies all the properties. Proof. We will prove this theorem when ∂D is smooth enough. For y ∈ D, x ∈ Rm \D the function |x − y|2−m is a harmonic function of y on D. Since |x − y|2−m → 0 as y → ∞ we can apply the Poisson formula (2.4.1.4) even if D is unbounded. Thus  ∂G 2−m = |x − y  |2−m (y, y  )dsy (2.5.3.1) |x − y| ∂n y ∂D where G is the Green function of D. From this we have    ∂G |x − y|2−m μ(dy) = |x − y  |2−m dsy (y, y  )μ(dy) . D ∂D D ∂ny   The inner integral is nonnegative, because ∂G ∂n > 0 for y ∈ ∂D. Let us denote  ∂G (y, y  )μ(dy) dsy . μb (dy  ) := D ∂n

Then we obtain the properties bal2) and bal3). The potential Π(x, μb ) is harmonic in D. Thus the function u(x) := Π(x, μb ) − Π(x, μ) is a subharmonic function (see Theorem 2.6.4.1). Every subharmonic function satisfies the maximum principle (see Theorem 2.6.1.2), i.e., u(x) < sup u(y) = 0. y∈∂D

Thus the property bal1) is fulfilled. To prove bal4) we can write the identity    ∂G  μb (dy ) = μ(dy) (y, y  )dy  . ∂n y ∂G G ∂G The inner integral is equal to 1 identically, because the function ≡ 1, y ∈ G is harmonic. Thus bal4) is true. Consider now the special case when m = 2, and D is an unbounded domain. Since log |1−z/ζ| → 0 when ζ → ∞, we obtain an equality like (2.5.3.1). Repeating the previous reasoning we obtain the last assertion for D with a smooth boundary. Exercise 2.5.3.1 Check this in detail. For the general case see [La, Ch. 4, § 1]; [Ca, Ch. 3, Thm. 4].



34

Chapter 2. Auxiliary Information. Subharmonic Functions

Pay attention that the swept potential Π(x, μb ) is also a solution of the Dirichlet problem in the domain D and the boundary function f (x) = Π(x, μ) in the following sense: Theorem 2.5.3.2 (Wiener) The equality bal2) holds in the points x ∈ ∂D which can be reached by the top of a cone placed outside D. For m = 2 it can fail only for isolated points. For the proof see [He], [La, Ch. 4, § 1, Thm. 4.3.]. The points of ∂D where the equality bal2) does not hold are called irregular. Theorem 2.5.3.3 (Kellogg’s Lemma) The set of all the irregular points of ∂D has zero capacity. For the proof see, e.g., [He], [La, Ch. 4, § 2, it. 10]. One can often compute the capacity using the following Theorem 2.5.3.4 (Equilibrium distribution) For any compact K with capm (K) > 0 there exists a mass distribution λK such that the following holds: eq1) Π(x, λ) = 1, x ∈ D\E, capm (E) = 0; eq2) supp λK ⊂ ∂K; eq3) λK (∂K) = capm (K). For the proof see [He], [La, Ch. 2, § 1, it. 3, Thm. 2.3]. Let us note that the set E in the previous theorem is a set of irregular points. The mass distribution λK is called equilibrium distribution, and the corresponding potential is called equilibrium potential. 2.5.4 Let h(x), x ≥ 0 be a positive continuous, monotonically increasing function which satisfies the condition h(0) = 0. Let {Kj} be a family of balls such that their diameters dj := d(Kj ) are no bigger then . Let us denote  1 mh (E, ) := inf h d(Kj ) , 2 where the infimum is taken over all coverings of the set E by the families {Kj }. The quantity mh (E) := lim mh (E, ) →0

is called h-Hausdorff measure [Ca, Ch. II]. Theorem 2.5.4.1 (Properties of mh ) The following properties hold: h1) monotonicity: E1 ⊂ E2 =⇒ mh (E1 ) ≤ mh (E2 ); h2) countable additivity:  mh (∪Ej ) = mh (Ej ); Ej ∩ Ei = ∅, for i = j; Ej ∈ σ(Rm ).

2.5. Potentials and capacities

35

We will quote two conditions (necessary and sufficient) that connect the hmeasure to the capacity (see, [La, Ch. 3, § 4, it. 9, 10]. Theorem 2.5.4.2 Let capE = 0. Then mh (E) = 0 for all h such that  h(r) dr < ∞. rm−1 0

Theorem 2.5.4.3 Let h(r) = rm−2 for m ≥ 3 and h(r) = (log 1/r)−1 for m = 2. If the h-measure of a set E is finite, then capm (E) = 0. Side by side with the Hausdorff measure the Carleson measure (see, [Ca, Ch. II], is often considered. It is defined by  mC h(0.5dj ), h (E) := inf where the infimum is taken over all coverings of the set E with balls of radii 0.5dj . The inequality mC h (E) ≤ mh (E) obviously holds. Let β − mesC E be the Carleson measure for h = rβ . The following assertion connects the β − mesC to capacity. Theorem 2.5.4.4 The following inequalities hold: β − mesC E ≤ N (m)(capm (E))β/m−2 , for m ≥ 3, β > m − 2; β − mesC (E) ≤ 18capl (E), for m = 2, β > 0, where N depends only on the dimension of the space. For the proof see [La, Ch. III,§ 4, it. 10, Cor. 2]. 2.5.5 Now we will formulate an analog of the Luzin theorem for potentials. Theorem 2.5.5.1 Let supp μ = K and let the potential Π(x, μ) be bounded on K. Then for any δ > 0 there exists a compact set K  ⊂ K such that μ(K\K  ) < δ and the potential Π(x, μ ) of the measure μ := μ |K (the restriction of μ to K) is continuous. For the proof see, e.g., [La, Ch. 3, § 2, it. 3, Thm. 3.6]. Let us prove the following assertion: Theorem 2.5.5.2 Let capK > 0. Then for arbitrarily small > 0 there exists a measure μ such that supp μ ⊂ K, the potential Π(x, μ) is continuous and μ(K) > cap(K) − . Proof. Consider the equilibrium distribution λK on K. Its potential is bounded by Theorem 2.5.3.4. By Theorem 2.5.5.1 we can find a mass distribution μ such that Π(x, μ) is continuous, supp μ ⊂ K and μ(K) > λK (K) − = cap(K) − . 

36

Chapter 2. Auxiliary Information. Subharmonic Functions

2.6 Subharmonic functions 2.6.1 Let u(x), x ∈ D ⊂ Rm be a measurable function bounded from above which can be −∞ on a set of no more than zero measure. Let us denote as  1 M(x, r, u) := u(y)dsy (2.6.1.1) σm rm−1 Sx,r the mean value of u(x) on the sphere Sx,r := {y : |y − x| = r}. The function M(x, r, u) is defined if Sx,r ⊂ D, but it can be −∞ a priori. A function u(x) is called subharmonic if it is upper semicontinuous, ≡ −∞, and for any x ∈ D there exists = (x) such that the inequality u(x) ≤ M(x, r, u)

(2.6.1.2)

holds for all r < . The class of functions subharmonic in D will be denoted as SH(D). Example 2.6.1.1 The function u(x) := −|x|2−m , x ∈ Rm belongs to SH(Rm ) for m ≥ 3, and the function u(z) := log |z|, z ∈ R2 is subharmonic in R2 . Example 2.6.1.2 Let f (z) be a holomorphic function in a plane domain D. Then log |f (z)| ∈ SH(D). Example 2.6.1.3 Let f = f (z1 , z2 , . . . , zn ) be a holomorphic function of z = (z1 , . . . , zn ). Then u(x1 , y1 , . . . , xn , yn ) := log |f (x1 + iy1 , . . . , xn + iyn )| is subharmonic in every pair (xj , yj ), and, as we can see later, in all the variables. Example 2.6.1.4 Every harmonic function is subharmonic, as follows from Theorem 2.4.1.6. (Mean Value). Theorem 2.6.1.1 (Elementary Properties) The following holds: sh1) if u ∈ SH(D), then Cu ∈ SH(D) for any constant C ≥ 0; sh2) if u1 , u2 ∈ SH(D), then u1 + u2 , max[u1 , u2 ] ∈ SH(D); sh3) suppose un ∈ SH(D), n = 1, 2, . . . , and the sequence converges to u monotonically decreasing or uniformly on every compact set in D. Then u ∈ SH(D);

2.6. Subharmonic functions

37

sh4) suppose u(x, y) ∈ SH(D1 ) for all y ∈ D2 , and be upper semicontinuous in D1 × D2. Let μ be a measure in D2 such that μ(D2 ) < ∞. Then the function u(x) := u(x, y)μ(dy) is subharmonic in D1 . sh5) let V ∈ SO(m) be an orthogonal transformation of the space Rm and u ∈ SH(Rm ). Then u(V •) ∈ SH(Rm ). All the assertions follow directly from the definition of subharmonic functions, properties of semicontinuous functions and properties of the Lebesgue integral. For a detailed proof see, e.g., [HK, Ch. 2]. Theorem 2.6.1.2 (Maximum Principle) Let u ∈ SH(D), G ⊂ Rm and u(x) ≡ const. Then the inequality u(x) < sup lim sup u(y), x ∈ D x ∈∂D y→x ,y∈D

holds. I.e., the maximum is not attained inside the domain. The assertion follows from (2.6.1.2) and the upper semicontinuity of u(x). For details see [HK, Ch. 2]. ◦

Let K  D be a compact set with nonempty interior K, and let fn be a decreasing sequence of functions continuous in K that tends to u ∈ SH(D). Such a sequence exists by Theorem 2.1.2.9. (The second criterion of semicontinuity). ◦

Consider a sequence {H(x, un )} of functions which are harmonic in K and H |∂K = fn . The sequence converges monotonically to a function H(x) harmonic ◦

in K by Theorem 2.3.4.3. (Connection between convergences), Theorem 2.4.1.8. (Uniform and D -convergences) and Theorem 2.6.1.2. The limit depends only on u as one can see, i.e., it does not depend on the sequence fn . This harmonic function H(x) := H(x, u, K) is called the least harmonic majorant of u in K. This name is justified because of the following Theorem 2.6.1.3. (Least Harmonic Majorant) Let u ∈ SH(D). Then for any K  ◦

D, u(x) ≤ H(x, u, K), x ∈ K. If h(x) is harmonic in K and satisfies the condition ◦

◦

h(x) ≥ u(x), x ∈ K, then H(x, u, K) ≤ h(x), x ∈ K. For the proof see [HK, Ch. 3]. 2.6.2 Let us study properties of the mean values of subharmonic functions. Let M(x, r.u) be defined by (2.6.1.1) and N (x, r, u) by  1 N (x, r, u) := u(y)dy, ωm rm Kx,r where ωm is the volume of the ball K0,1 .

38

Chapter 2. Auxiliary Information. Subharmonic Functions

Theorem 2.6.2.1 (Properties of Mean Values) The following holds: me1) M(x, r, u) and N (x, r, u) non-decreases in r monotonically; me2) u(x) ≤ N (x, •) ≤ M(x, •); me3) limr→0 M(x, r, u) = limr→0 N (x, r, u) = u(x). Proof. For simplicity let us prove me1) for m = 2. We have M(z0 , |z|, u) =

1 2π



2π

u(z0 + zeiφ )dφ. 0

Since u(z, φ) := u(z0 + zeiφ ) is a family of subharmonic functions that satisfies the condition sh4) of Theorem 2.6.1.1, M(z0 , |z|, u) is subharmonic in z on any K0,r . By Maximum Principle (Theorem 2.6.1.2) we have M(z0 , r1 , u) = max M(z0 , |z|, u) ≤ max M(z0 , |z|, u) = M(z0 , r2 , u) S0,r1

S0,r2

for r1 < r2 . Monotonicity of N (x, r, u) follows from the equality  N (x, r, u) = m

1

sm−1 M(x, rs.u)ds

(2.6.2.1)

0

and monotonicity of M(x, r, u). The property me2) follows now from the definition of a subharmonic function and (2.6.2.1). Let us prove me3). Let M (u, x, r) be defined by (2.1.1.1). We have M(x, r, u) ≤ M (u, x, r) and M (u, x, r) → u(x) because of upper semicontinuity of u(x). Thus me2) implies me3).



It is clear from me2) that a subharmonic function is locally summable. From me3) we have the corollary Theorem 2.6.2.2 (Uniqueness of subharmonic function) If u, v ∈ SH(D) and u = v almost everywhere, then u ≡ v. Let α(t) be defined by the equality (2.3.1.1), α (x) by (2.3.1.3). For a Borel set E let E  := {x : ∃y ∈ E : |x − y| < }. This is the -extension of E; this is, of course, an open set. For an open set D we set  D− := E . E  ⊂D

2.6. Subharmonic functions

39

This is the maximal set such that its -extension is a subset of D. One can see that D− is not empty for small and D− ↑ D when ↓ 0. Therefore for any D1  D there exists such that D1  D− . For u ∈ SH(D) set  (2.6.2.2) u (x) := u(x + y)α (y)dy which is defined in D− . Theorem 2.6.2.3 (Smooth Approximation) The following holds: ap1)

u is an infinitely differentiable subharmonic function in any open set D1 ⊂ D− ;

ap2)

u ↓ u(x) while ↓ 0 for all x ∈ D.

Proof. The property ap1) follows from sh4) (Theorem 2.6.1.1) and the following equality that one can obtain from (2.6.2.2):  (2.6.2.3) u (x) = u(y)α (x − y)dy. Exercise 2.6.2.1 Prove this. Let us prove ap2). From (2.6.2.2) we obtain  1 u (x) = α(s)sm−1 M(x, s, u)ds.

(2.6.2.4)

0

It follows from the property me1) (Theorem 2.6.2.1) that u1 ≤ u2 while 1 < 2 . Now we pass to the limit in (2.6.2.4). Using me3) we have M(x, s, u) ↓ u(x). We can pass to the limit under the integral because of Theorem 2.2.2.2. Thus  1 lim u (x) = α(s)sm−1 u(x)ds = u(x).  ↓0

0

Theorem 2.6.2.4 (Symmetry of u ) If u(x) depends only on |x| then u depends only on |x|. Proof. Let V ∈ SO(m) be a rotation of Rm . Then  u (V x) = u(y)α (V x − y)dy. Set y = V y  and change the variables. We obtain  u (V x) = u(V y  )α (V (x − y  ))dy. Since α = α (|x|) and u = u(|x|), α (V y) = α (y) and u(V y) = u(y). Thus  u (V x) = u (x) for any V and thus u (x) = u (|x|).

40

Chapter 2. Auxiliary Information. Subharmonic Functions

2.6.3 Since a subharmonic function is locally summable and defined uniquely by its values almost everywhere,every u ∈ SH(D) corresponds to a (unique) distribution  u, φ := u(x)φ(x)dx, φ ∈ D . Theorem 2.6.3.1 (Necessary Differential Condition for Subharmonicity) If u ∈ SH(D), then Δu is a positive distribution in D (D). Proof. Suppose to begin that u(x) has second continuous derivatives. By using (2.4.1.5) and (2.4.1.6) we can represent u(x) in the form  u(x) = M(x, r, u) + G(x, y, Kx,r )Δu(y)dy, (2.6.3.1) Kx,r

where G is negative for all r. Suppose Δu(x) < 0. Then it is negative in Kx,r for some r. Thus the integral in (2.6.3.1) is positive and we obtain that u(x) − M(x, r, u) > 0. This contradicts the subharmonicity of u(x). Now suppose u(x) is an arbitrarily subharmonic function. Then Δu (x) ≥ 0 for every x ∈ D when is small enough.For each x there is a neighborhood Dx such that every u defines a distribution from D (Dx ). Hence Δu (x) defines a positive distribution from D (Dx ). Passing to the limit in u when ↓ 0 we obtain in D (Dx ) a distribution that is defined by function u(x). Since the Laplace operator is continuous in any D (Theorem 2.3.4.2), Δu > 0 in D (Dx ). From Theorem  2.3.5.1 we obtain that Δu is a positive distribution in D (D). The distribution Δu can be realized as a measure by Theorem 2.3.2.2. The measure (θm )−1 Δu is called the Riesz measure of the subharmonic function u. Theorem 2.6.3.2 (Subharmonicity and Convexity) Let u(|x|) be subharmonic in x on K0,R . Then u(r) is convex with respect to −r2−m for m ≥ 3 and with respect to log r for m = 2. Proof. By Theorem 2.6.2.4, u (x) depends on |x| only, i.e., u (x) = u (|x|), and the function u (r) is smooth. Passing to the spherical coordinates we obtain Δu =

1 rm−1

∂ m−1 ∂ r u (r) ≥ 0. ∂r ∂r 1

By changing variables, r = ev for m = 2 or r = (−v) 2−m for m ≥ 3, we obtain [u (r(v)] ≥ 0, i.e., u (r(v)) is convex in v. Passing to the limit on ↓ 0 we obtain that u(r(v)) is convex too, as a monotonic limit of convex functions.  2.6.4 Now we will consider the connection between subharmonicity and potentials. Theorem 2.6.4.1 (Subharmonicity of −Π) −Π(x, μ, D) ∈ SH(D) It is because of GPo1) and GPo3) (Theorem 2.5.1.1).

2.6. Subharmonic functions

41

The following theorem is inverse to Theorem 2.6.3.1. Theorem 2.6.4.2 (Sufficient Differential Condition of Subharmonicity) Let Δu ∈ D (D) be a positive distribution. Then there exists u1 ∈ SH(D) that realizes u. −1 Δu. Let Ω1  Ω  D and Π(x, μΩ ) be the Newtonian (or Proof. Set μ := θm logarithmic) potential of μ |Ω . By GPo5) (Theorem 2.5.1.1) the sum H := u+Π is a harmonic distribution in D (Ω1 ). Hence there exists a “natural” harmonic function H1 that realizes H (Theorem 2.4.1.1). Thus the function u1 := H1 − Π ∈ SH(Ω1 ) and realizes u in D (Ω). Since Ω and Ω1 can be chosen such that a neighborhood of any x ∈ D belongs to Ω1 , the assertion holds for D. 

By the way, we showed in this theorem that every subharmonic function can be represented inside its domain of subharmonicity as a difference of a harmonic function and a Newton potential. Thus all the smooth properties of a subharmonic function depend on the smooth properties of the potential only because any harmonic function is infinitely differentiable. The following representation determines the harmonic function completely. Theorem 2.6.4.3 (F. Riesz representation) Let u ∈ SH(D) and let K be a compact Lipschitz subdomain of D. Then u(x) = H(x, u, K) − Π(x, μu , K) where μu is the Riesz measure of u and H(x, u, K) the least subharmonic majorant. Proof. We can prove as above that the function H(x) := u(x) + Π(x, μu , K) is ◦

harmonic in K. Since H(x) ≥ u(x) we have H(x) ≥ H(x, u, K). So we need the reverse inequality. Let us write the same equality for u that is smooth. u := H(x, u ) − Π(x, μu , K). Passing to the limit as ↓ 0 we obtain u(x) = H(x, u, K) − lim Π(x, μu , K), ↓0

and the potentials converge because other summands converge. By Gpo5) lim Π(x, μu , K) ≥ Π(x, μu , K). ↓0

Hence H(x) ≤ H(x, u, K).



2.6.5 In this item we will consider subharmonic functions in the ball KR := K0,R which are harmonic in some neighborhood of the origin and write u ∈ SH(R).

42

Chapter 2. Auxiliary Information. Subharmonic Functions

Set M (r, u) := max{u(x) : |x| = r}, μ(r, u) := μu (Kr ), M(r, u) := M(0, r, u),  r μ(t, u) N (r, u) := A(m) dt, where A(m) = max(1, m − 2). tm−1 0

(2.6.5.1)

Theorem 2.6.5.1 (Jensen-Privalov) For u ∈ SH(R), M(r, u) − u(0) = N (r, u), for 0 < r < R.

(2.6.5.2)

Proof. By Theorem 2.6.4.3 we have   1 r2 − |x|2 u(x) = u(y) dsy + G(x, y, Kr )μ(dy). σm r |y|=r |x − y|m Kr For x = 0 we obtain  r 1  1 μ(dt, u) + M(r, u), for m ≥ 3; − 0 tm−2 − rm−2 u(0) = r r − 0 log t μ(dt, u) + M(r, u), for m = 2. Integrating by parts gives   1  r r 1 −μ(t, u) tm−2 |0 +(m − 2) 0 − rm−2 u(0) − M(r, u) = r −μ(t, u) log rt |r0 + 0 μ(t,u) t dt,

μ(t,u) tm−1 dt,

for m ≥ 3;

for m = 2. (2.6.5.3) We have μ(t, u) = 0 for small t because of harmonicity of u(x). Thus (2.6.5.3) implies (2.6.5.2).  Theorem 2.6.5.2 (Convexity of M (r, u) and M(r, u)) These functions increase monotonically and are convex with respect to log r for m = 2 and −r2−m for m ≥ 3. Proof. Consider the case m = 2. Set M (z) := maxφ u(zeiφ ). One can see that M (r) = M (r, u). Let u be a continuous subharmonic function. Then M (z) is subharmonic (Theorem 2.6.1.1, sh5) and continuous because the family {uφ (z) := u(zeiφ )} is uniformly continuous. The function M (z) depends only on |z|. Thus it is convex with respect to log r by Theorem 2.6.3.2. Let u(z) be an arbitrarily subharmonic function and u ↓ u while ↓ 0. Then M (r, u ) ↓ M (r, u) by Proposition 2.1.2.7 and is convex with respect to log r by sh3), Theorem 2.6.1.1. If m ≥ 3 you should consider the function M (x) := max|y|=|x| u(Vy x) where Vy is a rotation of Rm transferring x into y. The convexity of M(r, u) is proved analogously.

2.7. Sequences of subharmonic functions

43

Exercise 2.6.5.1 Prove it. The monotonicity of M (r, u) follows from the Maximum Principle (Theorem 2.6.1.2). The monotonicity of M(r, u) was proved in Theorem 2.6.2.1.  The following classical assertion is a direct corollary of Theorem 2.6.5.2. Theorem 2.6.5.3 (Three Circles Theorem of Hadamard) Let f (z) be a holomorphic function in the disc KR and let Mf (r) be its maximum on the circle {|z| = r}. Then 1 Mf (r) ≤ ([Mf (r1 )]log

r2 r

r

[Mf (r2 )]log r1 )

log

r2 r1

for 0 < r1 ≤ r ≤ r2 < R. For the proof you should write down the condition of convexity with respect to log r of the function log Mf (r) which is the maximum of the subharmonic function log |f (z)| . Exercise 2.6.5.2 Do this. For details see [PS, Part I, Sec. III, Ch. 6, Problem 304].

2.7 Sequences of subharmonic functions 2.7.1 We will formulate the following analogue for the Montel theorem of normal families of holomorphic functions. The family (2.7.1.1) {uα , α ∈ A} ⊂ SH(D) is called precompact in D (D) if, for any sequence {αn , n = 1, 2, . . . } ⊂ A, there exists a subsequence αnj , j = 1, 2, . . . and a function u ∈ SH(D) such that uαnj → u in D (D). Example 2.7.1.1 uα := log |z − α|, |α| < 1 form a precompact family. Example 2.7.1.2 uα := log |fα | where {fα } is a family of holomorphic functions bounded in a domain D form a precompact family. A criterion of precompactness is given by Theorem 2.7.1.1 (Precompactness in D ) A family (2.7.1.1) is precompact iff the following conditions hold: comp1) for any compact set K ⊂ D a constant C(K) exists such that uα (x) ≤ C(K) for all α ∈ A and x ∈ K;

(2.7.1.2)

44

Chapter 2. Auxiliary Information. Subharmonic Functions

comp2) there exists a compact set K1  D such that inf max{uα (x) : x ∈ K1 } > −∞.

α∈A

(2.7.1.3)

For the proof see [H¨ o, Thm. 4.1.9]. Theorem 2.7.1.2 Let un → u in D (KR ). Then un → u in D (Sr ) for any r < R. Proof. We have μn → μ. Let us choose R1 such that r < R1 < R. Then un (x) = H(x, un , KR1 ) − Π(x, μn , KR1 ) by the F. Riesz theorem (Theorem 2.6.4.3). Now, we have Π(x, μn , KR1 ) → Π(x, μ, KR1 ) in D (R1 ) by GPo6), Theorem 2.5.1.1. Thus H(x, un , KR1 ) → H(x, u, KR1 ) in D (R1 ). By Theorem 2.4.1.8, H(x, un , KR1 ) → H(x, u, KR1 ) uniformly on any compact set in KR1 , in particular, on Sr . Hence H(x, un , KR1 ) → H(x, u, KR1 ) in D (Sr ). Also Π(x, μn , KR1 ) → Π(x, μ, KR1 ) in D (Sr ) by GPo6), Theorem 2.5.1.1.  Hence, un → u in D (Sr ). We say that a sequence fn of locally summable functions converges in Lloc to a locally summable function f if for any x ∈ D there exists a neighborhood  V  x such that V |fn − f |dx → 0. Theorem 2.7.1.3 (Compactness in Lloc ) Under conditions of Theorem 2.7.1.1 the family (2.7.1.1) is precompact in Lloc . For the proof see [H¨ o, Thm. 4.1.9]. +  Theorem 2.7.1.4 Let un → u in D (KR ). Then u+ n → u in D (KR ).

This is because u+ n (x) ≤ M, x ∈ K, for all compact sets K  KR . 2.7.2 The following theorem shows that a subharmonic function is much more “flexible” than a harmonic or analytic function. Theorem 2.7.2.1 Let D  Rm be a Lipschitz domain and let u ∈ SH(D) satisfy the condition u(x) < C for x ∈ D. Then for any closed domain D1  D there exists a function u ˜(x) := u ˜(x, D1 ) such that: ext1) ext2) ext3) ext4)

u(x) = u˜(x) f or x ∈ D1 ; u ˜(x) = C for x ∈ ∂D; u ˜ ∈ SH(D) and is harmonic in D\D1 ; u(x) ≤ u˜(x) f or x ∈ D.

The function u ˜ is defined uniquely.

2.7. Sequences of subharmonic functions

45

Proof. We can suppose without loss of generality that C = 0, because we can consider the function u − C. Let u(x) be continuous in D1 . Consider a harmonic function H(x) which is zero on ∂D and u(x) on ∂D1 . We have H(x) ≥ u(x) for x ∈ D\D1 because of Theorem 2.6.1.3. Set  H(x), x ∈ D\D1 ; u ˜(x) = u(x), x ∈ D1 . The function u ˜(x) is subharmonic in D. For x ∈ / ∂D1 it is obvious, and for x ∈ ∂D1 it follows from u(x) = u ˜(x) ≤ M(x, r, u) ≤ M(x, r, u ˜) for r small enough. It is easy to check that all the assertions of the theorem are fulfilled for the function u ˜. Exercise 2.7.1.1 Check this. Let u(x) be an arbitrarily subharmonic function. Consider the family u of smooth subharmonic functions that converges to u(x) decreasing monotonically  in a neighborhood of D1 . The sequence (u  ) converges monotonically to a subharmonic function that has all the properties ext1)–ext4).  ˜) Let un → u in D (D) and un (x) < 0 in D. Theorem 2.7.2.2 (Continuity of • Then for any K  D with a smooth boundary ∂K u n (•, K) → u ˜(•, K) in D (D). For proving, we need the following auxiliary statement: Theorem 2.7.2.3 Let un → u in D (D). Then for any smooth surface S  D and any function g(x) continuous in a neighborhood of S the assertion   un (x)g(x)dsx → u(x)g(x)dsx (2.7.2.1) S

S

holds. Proof. Since un → u in D (D) also the Riesz measures of the functions converge. Hence μn (K) ≤ C(K) for some K  S. Thus, for the sequence of potentials Π(x, μn ), we have    g(x)dsx Π(x, μn )g(x)dsx = μn (dy) . |x − y|m−2 S S The inner integral is a continuous function of y as can be seen by simple estimates. Thus the assertion (2.7.2.1) holds for potentials. Now, one can represent un in the form un (x) = Hn (x) − Π(x, μn ) in K. The sequence Hn convergences in D and, hence, uniformly on S. Thus  (2.7.2.1) holds for every un .

46

Chapter 2. Auxiliary Information. Subharmonic Functions ◦

Proof of Theorem 2.7.2.2. Let φ ∈ D(D) and supp φ ⊂ K. Then  un , φ = un , φ → u, φ = ˜ u, φ. Let x ∈ D\K. Then

 u n (x) = ∂K

∂G (x, y)un (y)dsy . ∂ny

n is precompact By Theorem 2.7.2.3, u n (x) → u˜(x) for x ∈ D\K. The sequence u ◦

in D (D). Thus every limit u0 of the u n coincides with u ˜(x) in K and in D\K. Hence, u0 ≡ u ˜ in D (D).  2.7.3 The property sh2), Theorem 2.6.1.1, shows that the maximum of any finite number of subharmonic functions is a subharmonic function too. However, it is not so if the number is not finite. Example 2.7.3.1 Set un (z) = n1 log |z|, n = 1, 2 . . . . The functions un ∈ SH(K1 ). Taking the supremum in n we obtain  0, for z = 0; u(z) =: sup un (z) = −∞ for z = 0. n The function is not semicontinuous, thus it is not subharmonic. However, it differs from a subharmonic function on a set of zero capacity. The following theorem shows that this holds in general. Theorem 2.7.3.1 (H. Cartan) Let a family {uα ∈ SH(D), α ∈ A} be bounded from above and u(x) := supα∈A uα (x). Then u∗ ∈ SH(D) and the set E := {x : u∗ (x) > u(x)} is a zero capacity set. For proving this theorem we need an auxiliary assertion Theorem 2.7.3.2 Let Π(x, μn , D) be a monotonically decreasing sequence of Green potentials and supp μn ⊂ K  D. Then there exists a measure μ such that the inequality lim Π(x, μn , D) ≥ Π(x, μ, D) n→∞

holds for all x ∈ D with equality outside some set of zero capacity. Proof. The sequence Π(x, μn , D) converges monotonically and thus in D (Theorem 2.3.4.3). Then μn → μ in D (Theorem 2.2.4.2.) and thus in C ∗ - topology (Theorem 2.3.4.4). By GPo5) (Theorem 2.5.1.1) we have lim Π(x, μn , D) ≥ Π(x, μ, D).

n→∞

Suppose that the strict inequality holds on some set E of a positive capacity. By Theorem 2.5.2.3 one can find a compact set K ⊂ E such that cap(K) > 0. Then there exists a measure ν concentrated on E such that its potential Π(x, ν, D) is

2.7. Sequences of subharmonic functions

47

continuous (Theorem 2.5.5.2). Thus we have    Π(x, μ, D)ν(dx) < lim Π(x, μn , D)ν(dx) = lim Π(x, μn , D)ν(dx), n→∞ n→∞    lim Π(x, ν, D)μn (dx) = Π(x, ν, D)μ(dx) = Π(x, μ, D)ν(dx). n→∞

The equalities use Theorem 2.2.2.2 (B. Levy), reciprocity law (GPo4), Theorem 2.5.1.1, C ∗ -convergence of μn and once more the reciprocity law, respectively. So we have a contradiction.  Proof of Theorem 2.7.3.1. Suppose that un (x) ↑ u(x). We can suppose also that ˜n (x) → u(x) for x ∈ G (see un < 0. For any domain G  D the sequence u Theorem 2.7.2.1), because un (x) = u ˜n (x) for x ∈ G. Since u˜n = Π(x, μ ˜n , D) for x ∈ D, u ˜(x) = Π(x, μ ˜ , D) = u(x) for x ∈ G and coincides with limn→∞ un (x) outside some set EG of zero capacity. Consider a sequence of domains Gn that exhaust D. Then u(x) = limn→∞ un (x) outside the set E := ∪∞ n=1 EGn which has zero capacity by capZ1) (see item 2.5.2). Now let {un , n = 1, 2 . . . } be a general countable set that satisfies the conditions of the theorem. Then the sequence vn := max{uk : k = 1, 2 . . . , n} ∈ SH(D) and vn ↑ u. Applying the previous reasoning we obtain the assertion of the theorem also in this case. Let {uα , α ∈ A} be an arbitrary set satisfying the condition of the theorem. By Theorem 2.1.3.2 (Choquet’s Lemma) one can find a countable set A0 ⊂ A such that (sup uα )∗ = (sup uα )∗ . A0

A

Since supA0 uα ≤ supA uα , we have     E := x : (sup uα )∗ > sup uα ⊂ E0 := x : (sup uα )∗ > sup uα . A

A

A0

A0



Thus cap (E) ≤cap (E0 ) = 0. Corollary of Theorem 2.7.3.1 is

Theorem 2.7.3.3 (H. Cartan +) Let {ut , t ∈ (0; ∞)} ⊂ SH(D) be a bounded from above family, and v := lim supt→∞ ut . Then v ∗ ∈ SH(D) and the set E := {x : v ∗ (x) > v(x)} has zero capacity. Proof. Set un := supt≥n ut , En := {x : (un )∗ > un }, E := ∪En . Since cap(En ) = 0, capE = 0 too. Let x ∈ / E. Then v(x) = lim sup ut (x) = lim (un )∗ (x). n→∞ t≥n

n→∞

48

Chapter 2. Auxiliary Information. Subharmonic Functions

The function

v ∗ := lim (un )∗ (x) n→∞

is the upper semicontinuous regularization of v(x) for all x ∈ D.



In spite of Example 2.7.3.1 we have Theorem 2.7.3.4 (Sigurdsson’s Lemma) [Si] Let S ⊂ SH(D) be compact in D . Then v(x) := sup{u(x) : u ∈ S} is upper semicontinuous and, hence, subharmonic. Proof. Note that u (x) = u, α(x − •) (see (2.6.2.3), (2.3.2.1)); and it is continuous in (u, x) with respect to the product topology on (SH(D) ∩ D ) × Rm (Theorem 2.3.4.6). Let x0 ∈ D, a ∈ R and assume that v(x0 ) < a. We have to prove that there exists a neighborhood X of x0 such that v(x) < a, x ∈ X.

(2.7.3.1)

We choose δ > 0 such that v(x0 ) < a−δ. If u0 ∈ SH(D) and is chosen sufficiently small, then u0 (x0 ) ≤ u0 (x0 ) < a − δ by Theorem 2.6.2.3 (Smooth Approximation). Since u (x) is continuous, there exists an open neighborhood U0 of u0 in SH(D) and an open neighborhood X0 of x0 such that u (x) < a − δ, u ∈ U0 , x ∈ X0 . The property ap2) (Theorem 2.6.2.3) implies u(x) < a − δ, u ∈ U0 , x ∈ X0 .

(2.7.3.2)

Since u0 is arbitrary and S is compact, there exists a finite covering U1 , U2 , . . . , Un of S and open neighborhoods X1 , X2 , . . . , Xn of x0 such that (2.7.3.2) holds for all  (u, x) : u ∈ Uj , x ∈ Xj , j = 1, . . . , n. Set X := ∩j Xj . Then (2.7.3.1) holds. 2.7.4 Now we are going to connect D -convergence to convergence outside a zero capacity set, the so-called quasi-everywhere convergence. Theorem 2.7.4.1 (D and Quasi-everywhere Convergence) Let un , u ∈ SH(D) and un → u in D (D). Then u(x) = lim supn→∞ un (x) quasi-everywhere and u(x) = (lim supn→∞ un (x))∗ everywhere in D.

2.7. Sequences of subharmonic functions

49

For the proof we need the following assertion in the spirit Theorem 2.7.3.2. Theorem 2.7.4.2 Let μn → μ in D (D) and supp μn ⊂ K  D. Then lim inf Π(x, μn , D) ≥ Π(μ, D) n→∞

with equality quasi-everywhere. Proof. The inequality was in GPo5), Theorem 2.5.1.1. Suppose the set E := {x : lim inf Π(x, μn , D) > Π(x, μ, D) n→∞

has a positive capacity. By Theorem 2.5.2.3 one can find a compact set K ⊂ E such that cap(K) > 0. By Theorem 2.5.5.2 one can find a measure ν concentrated on K with continuous potential. As in the proof of Theorem 2.7.3.2 we have    Π(x, μ, D)ν(dx) < lim inf Π(x, μn , D)ν(dx) ≤ lim inf Π(x, μn , D)ν(dx) n→∞ n→∞    = lim inf Π(x, ν, D)μn (dx) = Π(x, ν, D)μ(dx) = Π(x, μ, D)ν(dx). n→∞

The second inequality uses Theorem 2.2.2.3 (Fatou’s Lemma). The equalities use the reciprocity law (GPo4), Theorem 2.5.1.1, C ∗ -convergence of μn and once more the reciprocity law, respectively. So we have a contradiction.  Proof of Theorem 2.7.4.1. Let D1  D. Then the sequence un is bounded in D1 by Theorem 2.7.1.1. We can assume that un (x) < 0 for x ∈ D1 . ˜n (x, G) → u(x) in D (D1 ) by Theorem For any domain G  D1 the sequence u ˜n , D1 ). Thus μ ˜n → μ ˜ in D (D1 ). 2.7.2.2. We also have the equality u ˜n = −Π(x, μ By Theorem 2.7.4.2, lim inf n→∞ Π(x, μ ˜n , D1 ) = Π(x, μ ˜ , D1 ) quasi-everywhere in D1 . Hence (2.7.4.1) lim sup un = u n→∞

quasi-everywhere in G because un (x) = u ˜n (x) for x ∈ G. Consider a sequence of domains Gn that exhaust D. Then (2.7.4.1) holds outside a set En of zero capacity and (2.7.4.1) holds in D outside the set E := ∪∞ n=1 En which has zero capacity by capZ1) (see item 2.5.2), i.e., quasi-everywhere.  2.7.5 Now we connect the convergence of subharmonic functions in D to the convergence relative to the Carleson measure (see 2.5.4). We say that a sequence of functions un converges to a function u relative to the α-Carleson measure if the sets En := {x : |un (x) − u(x)| > } possess the property (2.7.5.1) α − mesC En → 0.

50

Chapter 2. Auxiliary Information. Subharmonic Functions

Theorem 2.7.5.1 (D and α-mesC Convergences) Let un , u ∈ SH(D) and un → u in D (D). Then for an every α > 0 and every domain G  D un → u relative to the (α + m − 2)-Carleson measure. For proving this theorem we need some auxiliary definitions and assertions. Let μ be a measure in Rm . We will call a point x ∈ Rm (α, α , )-normal with respect to the measure μ, (α < α ) if the inequality 

μx (t) := μ(Kx,t ) < −α tα+m−2 holds for all t < . Theorem 2.7.5.2 In any (α, α , )-normal point the following inequality holds:   [log |z − ζ| − log ]dμζ ≤ C α−α , for m = 2; − 

Kz, 

[|x − y|2−m − 2−m ]dμy ≤ C α−α , for m ≥ 3; Kx,t

while C = C(α, m) depends on α and m only. Proof. Let us consider the case m = 2. We have   

dμζ = log log dμz (t). |z − ζ| t 0 Kz, Integrating by parts we obtain    μz (t)

dμζ = log μz (t) |0 + dt log |z − ζ| t t 0 Kz,    1 −α ≤ tα−1 dt = α−α . α 0 Let us consider the case m ≥ 3. We have    2−m 2−m [|x − y| − ]dμy = (t2−m − 2−m )dμx (t) Kx,t

0

  μx (t) = (t2−m − 2−m )μx (t) |0 +(m − 2) dt m−2 0 t   m−2 m − 2 α−α ≤ α

tα−1 dt = . 

α 0

Theorem 2.7.5.3 (Ahlfors-Landkof Lemma) Let a set E ⊂ Rm be covered by balls with bounded radii such that every point is a center of a ball. Then there exists an at most countable subcovering of the same set with maximal multiplicity cr = cr(m). I.e., every point of E is covered no more than cr times. Let us note that cr(2) = 6. For the proof see [La, Ch. III, § 4, Lem. 3.2].

2.7. Sequences of subharmonic functions

51

Theorem 2.7.5.4 Let K  D. The set E := E(α, α , , μ) of points that belong to K and are not (α, α , )-normal with respect to μ satisfies the condition 

(α + m − 2) − mesC E ≤ cr(m) α μ(K  )

(2.7.5.2)

where K  is the 2 -extension of K. Proof. Let x ∈ E. Then there exists tx such that 

μx (tx ) ≥ tα+m−2

−α . x Thus every point of E is covered by a ball Kx,tx . By the Ahlfors-Landkof lemma (Theorem 2.7.5.3) one can find a no more than cr(m)-multiple subcovering {Kxj ,txj }. Then we have 



txα+m−2 ≤ cr(m) α μ(K  ). j

j

By definition of the Carleson measure we obtain (2.7.5.2).



Theorem 2.7.5.5 Let μn → μ in D (Rm ) and supp μn ⊂ K  Rm . Then for every α > 0 and G  Rm , Π(x, μn ) → Π(x, μ) relative to the (α + m − 2)-Carleson measure. Proof. Let m = 2. Set  log |z − ζ| =

log |z − ζ|,

for |z − ζ| > ,

log ,

for |z − ζ| ≤ .

This function is continuous for (z, ζ) ∈ K × K. Set νn := μn − μ. Then we have    − log |z − ζ|μn (dζ) + log |z − ζ|μ(dζ) = − log |z − ζ|νn (dζ)  =−

 log |z − ζ|νn (dζ) −

[log |z − ζ| − log ]νn (dζ). Kz,

The function log |z−ζ| is continuous in ζ uniformly over z ∈ K. Thus the sequence  Π (z) := log |z − ζ|νn (dζ) converges uniformly to zero on K. Suppose now that z∈ / E(α, α , , μ) ∪ E(α, α , , μn ),

52

Chapter 2. Auxiliary Information. Subharmonic Functions

i.e., it is an (α, α , )- normal point for μ and μn . By Theorem 2.7.5.2 we have   [log |z − ζ| − log ]νn (dζ) < 2C α−α . Kz,

Thus for sufficiently large n > n0 ( ),      |Π(z, μn ) − Π(z, μ)| =  log |z − ζ|μn (dζ) − log |z − ζ|μ(dζ) < δ = δ( ) while z ∈ / E(α, α , , μ) ∪ E(α, α , , μn ) := En ( ). By Theorem 2.7.5.3 the Carleson measure of En ( ) satisfies the inequality 



α − mesC En ( ) ≤ cr(m) α [μ(K) + μn (K)] ≤ C α := γ( ) where C = C(K) does not depend on n because μn (K) are bounded uniformly. Hence, for any > 0 the set En ( ) := {z : |Π(z, μn ) − Π(z, μ)| > δ( )} satisfies the condition

α − mesC En ( ) ≤ γ( )

(2.7.5.3)

while n > n0 = n0 ( ). Let us show that Π(z, μn ) → Π(z, μ) relative to α − mesC on K. Let γ0 , δ0 be arbitrarily small. One can find such that δ( ) < δ0 , γ( ) < γ0 . One can find n0 = n0 ( ) such that (2.7.5.3) is fulfilled. Now the set En,δ0 := {z : |Π(z, μn ) − Π(z, μ)| > δ0 } is contained in En ( ). Thus α − mesC En,δ0 < γ0 and this implies the convergence  relative to α − mesC . An analogous reasoning works for m ≥ 3. Proof of Theorem 2.7.5.1. Let un → u in D . One can assume that un , u are potentials on any compact set (Theorem 2.7.2.2). Hence, by Theorem 2.7.5.5 it  converges relative (α + m − 2) − mesC .

2.8 Scale of growth. Growth characteristics of subharmonic functions 2.8.1 Let A be a class of nondecreasing functions a(r), r ∈ (0, ∞) such that a(r) ≥ 0 and a(r) → ∞ when r → ∞. The quantity ρ[a] := lim sup r→∞

is called the order of a(r).

log a(r) log r

(2.8.1.1)

2.8. Scale of growth. Growth characteristics of subharmonic functions

53

Suppose ρ := ρ[a] < ∞. The number σ[a] := lim sup r→∞

a(r) rρ

(2.8.1.2)

is called the type number. If σ[a] = 0, we say a(r) has minimal type. If 0 < σ[a] < ∞, a(r) has normal type. If σ[a] = ∞, it has maximal type. Example 2.8.1.1 Set a(r) := σ0 rρ0 . Then ρ[a] = ρ0 , σ[a] = σ0 . Example 2.8.1.2 Set a(r) := (log r)−1 rρ0 . Then ρ[a] = ρ0 , σ[a] = 0. Example 2.8.1.3 Set a(r) := (log r)rρ0 . Then ρ[a] = ρ0 , σ[a] = ∞. Theorem 2.8.1.1 (Convergence Exponent) The following equality holds:    ∞ a(r)dr <∞ . (2.8.1.3) ρ[a] = inf λ : rλ+1 If the integral converges for λ = ρ[a], a(r) has minimal type. Exercise 2.8.1.1 Prove this. For the proof see, e.g., [HK, § 4.2]. Example 2.8.1.4 Let rj , j = 1, 2, . . . be a nondecreasing sequence of positive numbers. Let us concentrate the unit mass in every point rj and define a mass distribution n(E) := {the number points of the sequence {rj } in E}, E ⊂ R. Then

 0

∞

∞

dn  1 = . rλ rjλ 1

(2.8.1.4)

The infimum of λ for which the series in (2.8.1.4) converges is usually called the convergence exponent for the sequence {rj } [PS, Part I, Sec. 1, Ch. III, § 2]. Integrating by parts one can transform the integral in (2.8.1.4) to an integral of the form (2.8.1.3) where a(r) = n((−∞, r)). Theorem 2.8.1.1 shows that the convergence exponent coincides with the order of this a(r). A function ρ(r) is called a proximate order with respect to order ρ if po1)

ρ(r) ≥ 0,

po2)

limr→∞ ρ(r) = ρ,

po3)

ρ(r) has a continuous derivative on (0, ∞),

po4)

limr→∞ r log rρ (r) = 0.

54

Chapter 2. Auxiliary Information. Subharmonic Functions

Two proximate orders ρ1 (r) and ρ2 (r) are called equivalent, if  1 ρ1 (r) − ρ2 (r) = o . log r

(2.8.1.5)

For a ∈ A set

a(r) . (2.8.1.6) rρ(r) It is called a type number with respect to a proximate order ρ(r). It is clear that this type number is the same for equivalent proximate orders. σ[a, ρ(r)] := lim sup r→∞

Theorem 2.8.1.2 (Proper Proximate Order) Let a ∈ A and ρ[a] = ρ < ∞. Then there exists a proximate order ρ(r) such that 0 < σ[a, ρ(r)] < ∞.

(2.8.1.7)

For the proof see [Le, Ch. 1, Sec. 12, Thm. 16]. If a proximate order satisfies the condition (2.8.1.7), we will call it the proper proximate order of a(r) (p.p.o.). The function rρ(r) inherits a lot of useful properties of the power function rρ . Theorem 2.8.1.3 (Properties of P.O) The following holds: ppo1) the function V (r) := rρ(r) increases monotonically for sufficiently large values of r. ppo2) for q < ρ + 1,  r rρ(r)+1−q tρ(t)−q dt ∼ ρ+1−q 1 and for q > ρ + 1,  ∞ rρ(r)+1−q tρ(t)−q dt ∼ q−ρ−1 r as r → ∞. ppo3) the function L(r) := rρ(r)−ρ satisfies the condition ∀δ > 0, L(kr)/L(r) → 1 when r → ∞ uniformly for k ∈ [ 1δ , δ]. Exercise 2.8.1.2 Prove these properties. For the proof see, e.g., [Le, Ch. 2, Sec. 12]. The following assertion allows us to replace any p.o. with a smooth one. Theorem 2.8.1.4 (Smooth P.O) Let ρ(r) be an arbitrary p.o. There exists an infinitely differentiable equivalent p.o. ρ1 (r) such that (k)

rk log rρ1 (r) → 0, k = 1, 2, . . . when r → ∞.

(2.8.1.8)

2.8. Scale of growth. Growth characteristics of subharmonic functions

55

Proof. Let α be defined by (2.3.1.3). Set := 0.5, po(x) := ρ(ex ) and 

x

po1 (x) := po(n) + [po(n + 1) − po(n)]

α0.5 (t + 0.5)dt n

for x ∈ [n, n + 1). The function po1 (x) is continuous and infinitely differentiable due to properties of α and po1 (n) = po(n) for n = 1, 2, . . . . By property po3) of p.o. we have (n + 1)|po(n + 1) − po(n)| ≤

n+1 max |y · po (y)| → 0 n y∈[n,n+1]

as n → ∞. Thus max y∈[n,n+1]

(k)

|y · po1 (y)| ≤ const ·(n + 1)|po(n + 1) − po(n)| → 0

as n → ∞. So ρ1 (r) := po1 (log r) is a p.o. that satisfies (2.8.1.8). Let us show that it is equivalent to ρ(r). Indeed   x    dy   [po(y) − po1 (y)] y  |po(x) − po1 (x)| =  y n

≤

max [|y · po (y)| + |y · po1 (y)|] log

y∈[n,n+1]

when x ∈ [n, n + 1] and n → ∞.

n+1 =o n

 1 , x 

We will further need (in 2.9.3) the following assertion. Theorem 2.8.1.5. (A.A. Gol dberg) Let ρ(r) → ρ be a p.o., and let f (t) be a function that is locally summable on (0, ∞) and such that lim tρ+δ f (t) = lim tρ+1+γ f (t) = 0 t→∞

t→0

(2.8.1.9)

for some 0 < δ, γ < 1. Then lim r−ρ(r)



r→∞

lim r

r→∞

−ρ(r)



x

x

(rt)ρ(rt) f (t)dt = −1 cr ∞

tρ f (t)dt, 

0

ρ(rt)

(rt) x

for any c > 0 and any x ∈ (0, ∞).

∞

f (t)dt =

(2.8.1.10) ρ

t f (t)dt x

56

Chapter 2. Auxiliary Information. Subharmonic Functions

Proof. Set



∞

I(r) := cr −1

(rt)ρ(rt) f (t)dt. rρ(r)

It will be enough to prove that 

∞

lim I(r) =

r→∞

tρ f (t)dt

(2.8.1.11)

0

because both functions  f0 (t, x) :=

for t ∈ (0, x), for t ∈ [x, ∞)

f (t), 0

and f∞ (t, x) := f (t) − f0 (t, x) also satisfy the condition of the theorem. Let us represent the integral as the following sum: 

∞

I(r) := cr −1

(rt)ρ(rt) f (t)dt = I1 (r, ) + I2 (r, ) + I3 (r, ), rρ(r)

where

 I1 (r, ) :=



cr −1



−1

I2 (r, ) := 

 I3 (r, ) :=

∞

−1

(2.8.1.12)

(rt)ρ(rt) f (t)dt, rρ(r) (rt)ρ(rt) f (t)dt, rρ(r) (rt)ρ(rt) f (t)dt. rρ(r)

We can represent I2 (r, ) in the form 

−1

I2 (r, ) = 

L(rt) ρ t f (t)dt. L(r)

By ppo3) (Theorem 2.8.1.3), 

−1

tρ f (t)dt.

lim I2 (r, ) =

r→∞



Let us estimate the “tails”. From (2.8.1.9) we have |f (t)| ≤ Ct−ρ−δ for 0 < t ≤ where C does not depend on and |f (t)| ≤ Ct−ρ−1−γ

(2.8.1.13)

2.8. Scale of growth. Growth characteristics of subharmonic functions

for t ≥ −1 . We have



|I1 (r, )| ≤ C



cr −1

57

(rt)ρ(rt) −ρ−δ t dt := CJ1 (r, ) rρ(r)

and lim sup |I1 (r, )| ≤ C lim J1 (r, ). r→∞

r→∞

(2.8.1.14)

Let us calculate the last limit.We perform the change x = tr:  r t−ρ(x)−(ρ+δ) dx. J1 (r, ) = r−ρ(r)+ρ+δ−1 c

Now we use ppo2) for q = ρ + δ and ppo3): ( r)ρ(r)−(ρ+δ)+1 1 lim 1 − δ r→∞ rρ(r)−(ρ+δ)+1 L( r)

1−δ

1−δ lim = . = 1 − δ r→∞ L(r) 1−δ

lim J1 (r, ) =

r→∞

Substituting in (2.8.1.14) we obtain lim sup |I1 (r, )| ≤ C r→∞

1−δ . 1−δ

(2.8.1.15)

γ . γ

(2.8.1.16)

Analogously one can obtain lim sup |I3 (r, )| ≤ C r→∞

Using (2.8.1.13), (2.8.1.15) and (2.8.1.16) one can pass to the limit in (2.8.1.12) as r → ∞, then let → 0 and obtain (2.8.1.11).  2.8.2 Let u(x) := u1 (x) − u2 (x)

(2.8.2.1)

where u1 , u2 ∈ SH(R ), u1 (0) > −∞, u2 (0) = 0 and μ1 := μu1 , μ2 := μu2 are concentrated on disjoint sets. Let m = 2, uj (z) := log |fj (z)|, j = 1, 2 where fj (z), j = 1, 2 are entire functions. Then the function u(z) = log |f (z)|, where f (z) := f1 (z)/f2 (z), is meromorphic. The condition for masses means that f1 and f2 have no common zeros, u2 (0) = 0 corresponds to f2 (0) = 1 and u1 (0) > −∞ means f1 (0) = 0. The class of such functions is denoted as δSH(Rm ). In spite of the standardization conditions the representation (2.8.2.1) is not unique. However for any pair of representations u1 − u2 and u1 − u2 , m

uj (x) − uj (x) = Hj (x), j = 1, 2 where Hj are harmonic and H2 (0) = 0.

(2.8.2.2)

58

Chapter 2. Auxiliary Information. Subharmonic Functions

Really, from the equality u1 − u2 = u1 − u2 we obtain μ1 − μ2 = μ1 − μ2 . Using Theorem 2.2.1.2 (Jordan decomposition) we obtain μ1 = μ1 , μ2 = μ2 . Thus (2.8.2.2) holds. Obviously H2 (0) = 0. Set  1 max(u1 , u2 )(ry)dy (2.8.2.3) T (r, u) := σm |y|=1 where σm is the surface square of the unit sphere. It is called the Nevanlinna characteristic of u ∈ δSH(Rm ). The Nevanlinna characteristic does not depend on the representation (2.8.2.1). Indeed,   max(u1 , u2 )(ry)dy = [(u1 − u2 )+ (ry) − u2 (ry)]dy |y|=1 |y|=1  = [(u1 − u2 )+ (ry) − u2 (ry) + H2 (rx)]dy |y|=1  = [max(u1 , u2 )(ry) + H2 (rx)]dy |y|=1  max(u1 , u2 )(ry)dy + H2 (0) = |y|=1  = max(u1 , u2 )(ry)dy. |y|=1

Note also that the class δSH(Rm ) is linear. Actually, let u ∈ δSH(Rm ). Then λu ∈ δSH(Rm ) for λ > 0. The function −u ∈ δSH(Rm ), since −u(x) = [u2 (x) − u1 (0)] − [u1 (x) − u1 (0)]. Let us show that u1 + u2 ∈ δSH(Rm ) if u, v ∈ δSH(Rm ). Set ν := νu + νv , where νu , νv are the corresponding charges. By Theorem 2.2.1.2 (Jordan decomposition) ν = ν + − ν − , where ν + , ν − are measures concentrated on disjoint sets. Let u1 be a subharmonic function in Rm the mass distribution of which coincides with ν + .2 Then u2 := u1 − (u + v) is a subharmonic function with the mass distribution ν − . Hence u(x) + v(x) = [u1 (x) − u2 (0)] − [u2 (x) − u2 (0)]. Theorem 2.8.2.1 (Properties T (r, u)) The following holds: t1) T (r, u) increases monotonically and is convex with respect to −rm−2 for m = 2 and with respect to log r for m = 2. 2 We

will give the construction of such a function for the case of finite order (item 2.9.2), but it is possible actually always, see, for example, [HK, Thm. 4.1]

2.8. Scale of growth. Growth characteristics of subharmonic functions

59

t2) For u ∈ SH(Rm ), (i.e., u2 ≡ 0) T (r, u) =

1 σm

 u+ (ry)dy. |y|=1

t3) T (r, u) = T (r, −u) − u1 (0). t4) T (r, u + u ) ≤ T (r, u) + T (r, u ), T (r, λu) = λT (r, u) for λ > 0. Proof. Since v(x) := max(u1 , u2 )(x) is subharmonic, t1) follows from Theorem 2.6.5.2 (Convexity of M (r, u) and M(r, u)). The property t2) is obvious, t3) follows from the equality −u(x) = u2 (x) − [u1 (x) − u1 (0)] − u1 (0). The properties t4) follow from the properties of maximum and t3).  Set ρT [u] := ρ[a] (see, (2.8.1.1)) where a(r) := T (r, u). It is called the order of u(x) with respect to T (r). Theorem 2.8.2.2 (ρT -property) For u1 , u2 ∈ δSH(Rm ) the following inequality holds: ρT [u1 + u2 ] ≤ max(ρT [u1 ], ρT [u2 ]). (2.8.2.4) Equality in (2.8.2.4) is attained if ρT [u1 ] = ρT [u2 ]. Proof. Set u := u1 + u2 . From t3) and t4) T (r, u) ≤ T (r, u1 ) + T (r, u2 ) + O(1) ≤ 2 max[T (r, u1 ), T (r, u2 )] + O(1). From the definition of ρT we obtain (2.8.2.4). Suppose, for example, ρT [u1 ] > ρT [u2 ]. Let us show that ρT [u] = ρT [u1 ]. From the equality u1 = u + (−u2) we obtain ρT [u1 ] ≤ max(ρT [u], ρT [u2 ] If ρT [u] < ρT [u1 ], then from the previous inequality we would have the contradiction ρT [u1 ] < ρT [u1 ].  Let us define σT [u] by (2.8.1.2) while ρ := ρT [u]. Set also σT [u, ρ(r)] := σ[a, ρ(r)] (see (2.8.1.6)), where a(r) := T (r, u). The characteristics ρT [u], σT [u], σT [u, ρ(r)] are defined for u ∈ δSH(Rm ). For the class of subharmonic functions we have the inclusion SH(Rm ) ⊂ δSH(Rm ) and, of course, all these characteristics can be applied to a subharmonic function. However, for the class SH(Rm ) the standard characteristic of growth is M (r, u) that we can not apply to a δ-subharmonic function u ∈ δSH(Rm ). Thus for u ∈ SH(Rm ) we define new characteristics ρM [u], σM [u], σM [u, ρ(r)] in the same way by replacing T (r, u) for M (r, u). The following theorem shows that there is not a big difference between characteristics with respect to T and M for u ∈ SH(Rm ).

60

Chapter 2. Auxiliary Information. Subharmonic Functions

Theorem 2.8.2.3 (T and M -characteristics) Let u ∈ SH(Rm ) and ρ(r)(→ ρ) any p.o. Then ρMT1) ρMT2)

ρT [u] and ρM [u] are finite simultaneously and ρT [u] = ρM [u] := ρ[u] there exists A := A(ρ, m) such that AσM [u, ρ(r)] ≤ σT [u, ρ(r)] ≤ σM [u, ρ(r)].

In particular, the last property means that the types with respect to T (r) and M (r) for the same p.o. are minimal, normal or maximal at the same time. Proof. From t2), Theorem 2.8.2.1 we have T (r, u) ≤ M (r, u) for u ∈ SH(Rm ). Thus ρT [u] ≤ ρM [u], proving the second part of ρMT2). Let H(x) be the least harmonic majorant of u(x) in the ball K2R . By the Poisson formula (Theorem 2.4.1.5) and Theorem 2.6.1.3,  1 (4R2 − |x|2 ) u(y) dsy M (R, u) ≤ M (R, H) = max |x − y|m |x|=R σm 2R |y|=2R  2m−2 ≤ |u(2Ry)|dsy (2.8.2.5) σm |y|=1 = 2m−2 [T (2R, u) + T (2R, −u)] = 2m−2 [2T (2R, u) − u(0)]. From here one can obtain ρT [u] ≥ ρM [u] . The left side of ρMT2) with A(ρ, m) :=  2−ρ−m+2 follows from the properties of p.o. Exercise 2.8.2.1 Prove the first inequality from ρMT2). 2.8.3 Let μ be a mass distribution (measure) in Rm (μ ∈ M(Rm )). The characteristic ρ[μ] := ρ[a] − m + 2 for a(r) := μ(Kr ) (see (2.8.1.1)) is called the convergence exponent of μ, and ¯ Δ[μ] := σ[a] for the same a (see (2.8.1.2)) is called the upper density of μ. The least integer number p for which the integral  ∞ μ(t) dt tp+m

(2.8.3.1)

converges is called the genus of μ and is denoted p[μ]. Theorem 2.8.3.1 (Convergence Exponent and Genus) The following holds: ceg1) ceg2)

p[μ] ≤ ρ[μ] ≤ p[μ] + 1, ¯ for ρ[μ] = p[μ] + 1, Δ[μ] = 0.

2.8. Scale of growth. Growth characteristics of subharmonic functions

61

Proof. From Theorem 2.8.1.1 (Convergence Exponent) we have ρ[μ] + 1 + m − 2 ≤ p[μ] + m. Thus ρ[μ] ≤ p[μ] + 1. The same theorem implies ρ[μ] + m − 2 + 1 ≥ p[μ] + m − 1. Thus p[μ] ≤ ρ[μ], and ceg1) is proved. Let ρ(μ) = p[μ] + 1. Then the integral (2.8.3.1) converges for p[μ] = ρ[μ] − 1. We use the inequality  ∞  ∞ μ(t) dt μ(r) dt ≥ μ(r) dt = ρ[μ]+m−2 (ρ[μ] + m − 2)−1 . ρ[μ]+m−1 ρ[μ]+m−1 t t r r r Since the left side of the inequality tends to zero we obtain ¯ Δ[μ] = lim

r→∞

μ(r) rρ[μ]+m−2



= 0.

Set ¯ ρ(r)] := σ[a, ρ(r) + m − 2], Δ[μ,

(2.8.3.2)

where a(r) := μ(r) (see (2.8.1.6)). It is clear that ρ(r) + m − 2 is also a p.o. Set as in (2.6.5.1),  r μ(t) N (r, μ) := A(m) dt, m−1 t 0 where A(m) = max(1, m − 2). Set also ¯ N [μ, ρ(r)] := σ[a, ρ(r)], ρN [μ] := ρ[a], Δ where a(r) := N (r, μ). This is the N-order of μ and the N-type of μ with respect to p.o. ρ(r). Theorem 2.8.3.2 (N -order and Convergence Exponent) The following holds: Nce1) ρN [μ] and ρ[μ] are finite simultaneously and ρN [μ] = ρ[μ], Nce2) for ρ > 0 there exists such Aj := Aj (ρ, m), j = 1, 2, that ¯ ρ(r)] ≤ Δ ¯ N [μ, ρ(r)] ≤ A2 Δ[μ, ¯ ρ(r)]. A1 Δ[μ, Proof. We have the inequality  2r  2r μ(t) dt μ(r) N (2r, μ) ≥ A(m) dt ≥ A(m)μ(r) ≥ A(m)B(m) , m−1 m−1 t t (2r)m−2 r r where B(m) := 1 − 22−m for m ≥ 3 and B(2) := log 2. From here one can obtain the inequality ρ[μ] ≥ ρN [μ] and the left side of Nce2) for A1 (ρ, m) := A(m)B(m)2−ρ . For proving the opposite inequalities we use the l’Hˆospital Rule (slightly improved): lim sup r→∞

1¯ N (r, μ) N  (r, μ) μ(r)r2−m = Δ[μ]. ≤ lim sup ρ(r)  = lim sup ρ(r) ρ(r)  ρ r ) [ρ(r) + r log rρ (r)] r→∞ (r r→∞ r

Thus ρN [μ] ≤ ρ[μ] and the right side of Nce2) holds.



62

Chapter 2. Auxiliary Information. Subharmonic Functions

We shall denote as δM(Rm ) the set of charges (signed measures) of the form ν := μ1 − μ2 where μ1 , μ2 ∈ M(Rm ). Let us remember that |ν| ∈ M(Rm ) is the full variation of ν (see 2.2.1). Theorem 2.8.3.3 (Jensen) Let u := u1 − u2 ∈ δSH(Rm ) and ν := μ1 − μ2 be a corresponding charge. Then J1) ρ[|ν|] ≤ max(ρ[μ1 ], ρ[μ2 ]) ≤ ρ[u], ¯ ¯ 1 , ρ(r)]+ Δ[μ ¯ 2 , ρ(r)] ≤ AσT [u, ρ(r)] for some A := A(ρ, m). J2) Δ[|ν|, ρ(r)] ≤ Δ[μ Proof. We can suppose without loss of generality that u(0) = 0 because the function u(x) − u(0) has the same order and the same number type if ρ > 0. We apply the Jensen-Privalov formula (Theorem 2.6.5.1) to the functions u1 , u2 and obtain N (r, μj ) ≤ M(r, uj ) ≤ T (r, u). Thus N (r, |ν|) ≤ N (r, μ1 )+ N (r, μ2 ) ≤ 2T (r, u). From here one can obtain J1) and ¯ N [|ν|, ρ(r)]. However, we can delete the subscript N because J2) for ρN [|ν|] and Δ of Theorem 2.8.3.2. 

2.9 The representation theorem of subharmonic functions in Rm 2.9.1 Set  H(z, cos γ, m) :=

− 12 log(z 2 − 2z cos γ + 1), − m−2 2

(z 2 − 2z cos γ + 1)

,

for m = 2, for m ≥ 3.

(2.9.1.1)

The function H(z, cos γ, m) is holomorphic on z in the disk {|z| < 1}. It can be represented there in the form H(z, cos γ, m) =

∞ 

m−2 2

Ck

(cos γ)z k

(2.9.1.2)

k=0

where every coefficient Ckβ (•), k = 0, 1, . . . is a polynomial of degree k.

1

Such polynomials are called the Gegenbauer polynomials. Note that Ck2 (•) are the Legendre polynomials and  0, for k = 0, 0 Ck (λ) = 1 for k ≥ 1, k cos(k arccos λ), i.e., they are proportional to the Chebyshev polynomials.

2.9. The representation theorem of subharmonic functions in Rm

63

Thus for m = 2 we have the equality ∞

 cos kγ 1 zk − log(z 2 − 2z cos γ + 1) = 2 k k=1

that can be checked directly. Let x ∈ Rm . Set x0 := x/|x|. Then the scalar product (x0 , y 0 ) is equal to cos γ where γ is the angle between x and y. Let Em (x) be defined by (2.4.1.1). For m ≥ 3 the function Em (x − y) is the Green function for Rm . One can see that it is represented in the form G(x, y, Rm ) := Em (x − y) = −|y|2−m H(|x|/|y|, cos γ, m) where cos γ = (x0 , y 0 ). For m = 2 the function −H(|x|/|y|, cos γ, 2) plays the same role. Thus we will denote it as G(x, y, R2 ). Theorem 2.9.1.1 (Expansion of G(x, y, Rm )) The following holds: G(x, y, R ) = − m

∞ 

m−2 2

Ck

(cos γ)

k=0

|x|k |y|k+m−2

,

(2.9.1.3)

for |x| < |y|, and the functions m−2 2

Dk (x, y) := Ck

(cos γ)

|x|k |y|k+m−2

(2.9.1.4)

are homogeneous harmonic functions in x and harmonic in y for y = 0. Proof. The expansion (2.9.1.3) follows from (2.9.1.2). The function G(zx, y, Rm ) is harmonic for |x| < |y| and, hence, for any real 0 ≤ z < 1. Hence, for any ψ ∈ D(Kr ) while r := 0.5|y| the function g(z) := G(z•, y, Rm ), Δψ = 0 for z ∈ (0, 1). The function g is holomorphic for all complex z ∈ {|z| < 1} because G(zx, y, Rm ) is holomorphic. Thus g(z) ≡ 0, i.e., all its coefficients are zero. From the expansion (2.9.1.3) we can see that the coefficients of G(zx, y, Rm ) are Dk (x, y). Hence, Dk (•, y), Δψ = 0 for every ψ ∈ D(Kr ). Thus Dk (•, y) is a harmonic distribution. By Theorem 2.4.1.1 it is an ordinary harmonic function for |x| < 0.5|y|. m−2

Ck 2 (cos γ) is a polynomial of degree k with respect to (x0 , y 0 ). Thus Dk (x, y) is a homogeneous polynomial of x and is harmonic for all x. Let us prove the harmonicity in y. By Theorem 2.4.1.10 the function Dk (y ∗ , x0 )|y|2−m (∗ stands for inversion) is harmonic in y. We have Dk (y ∗ , x0 )|y|2−m = |y|2−m Dk (y/|y|2 , x0 ) = Dk (x0 , y).



64

Chapter 2. Auxiliary Information. Subharmonic Functions

Set H(z, cos γ, m, p) = H(z, cos γ, m) −

p 

m−2 2

Ck

(cos γ)z k .

(2.9.1.5)

k=0

Theorem 2.9.1.2 The following holds:

for |z| ≤ 1/2, and

|H(z, cos γ, m, p)| ≤ A1 (m, p)|z|p+1

(2.9.1.6)

|H(z, cos γ, m, p)| ≤ A2 (m, p)|z|p

(2.9.1.7)

for |z| ≥ 2, −π < arg z ≤ π. The factor |z|p should be replaced by log |z| if m = 2, p = 0. Proof. Consider the function φ(z) := H(z, cos γ, m, p)z −p−1 . It is holomorphic in the disk {|z| ≤ 1/2}. We apply the maximum principle and obtain (2.9.1.6) where A1 (m, p) = 2p+1 max |φ(z)|. |z|=1/2

For proving (2.9.1.7) we consider the function ψ(z) := H(z, cos γ, m, p)z −p that is holomorphic in the domain D := {z : |z| ≥ 2, −π < arg z ≤ π} and continuous in its closure. Applying the maximum principle we obtain (2.9.1.7) where A2 (m, p) = 2p max |ψ(z)|. z∈∂D



Set Gp (x, y, m) := −|y|2−m H(|x|/|y|, cos γ, m, p) where cos γ = (x0 , y 0 ). Note the equality Gp (x, y, m) = G(x, y, Rm ) +

p 

Dk (x, y).

k=0

Exercise 2.9.1.1 Check this using (2.9.1.3), (2.9.1.4) and (2.9.1.5). It looks like a Green function for Rm but it tends more quickly to zero at infinity and generally speaking it is not negative. For m = 2 it can be represented in the form Gp (z, ζ, 2) = log |E(z/ζ, p)| where E(z/ζ, p) is the primary Weierstrass factor:    2  p ! z 1 z z 1 z E(z/ζ, p) := 1 − +···+ . exp + ζ ζ 2 ζ p ζ We will call it the primary kernel analogously to the primary factor.

2.9. The representation theorem of subharmonic functions in Rm

65

Theorem 2.9.1.3 (Estimate of Primary Kernel) The following holds: |Gp (x, y, m)| ≤ A(m, p) for |x| < 2|y|, |Gp (x, y, m)| ≤ A(m, p)

|x|p+1 |y|p+m−1

(2.9.1.8)

|x|p

(2.9.1.9)

|y|p+m−2

for |y| < 2|x|, and  Gp (x, y, m) ≤ A(m, p) min

|x|p+1 |x|p , p+m−2 p+m−1 |y| |y|

(2.9.1.10)

for all x, y ∈ Rm , where A(m, p) does not depend on x, y. For m = 2, p = 0 we have Gp (z, ζ, 2) ≤ A(2, 0) log(1 +

|z| |ζ| ).

Proof. The inequality (2.9.1.8) follows directly from (2.9.1.6) and (2.9.1.9) follows from (2.9.1.7). By the condition 2 ≤ |x|/|y| (2.9.1.10) follows from (2.9.1.9). Suppose 1/2 ≤ |x|/|y| ≤ 2. Since all the summands in (2.9.1.5) are bounded from below, for 1/2 ≤ z ≤ 2 we have  Gp (x, y, m) ≤ A1 (m, p)|y|

2−m

≤ A(m, p) min

also under these conditions. The case m = 2, p = 0 is obvious.

|x|p+1 |x|p , |y|p+m−1 |y|p+m−2





2.9.2 Let μ ∈ M(Rm ). We suppose below that its support does not contain the origin.  We will say that the integral Rm f (x, y)μ(dy) converges uniformly on x ∈ D if       f (x, y)μ(dy) → 0 sup    x∈D |y|>R when R → ∞. Hence, the integral is permitted to be equal to infinity for some finite x. Let μ have genus p (see, 2.8.3). Set  Gp (x, y, m)μ(dy). (2.9.2.1) Π(x, μ, p) := Rm

It is called the canonical potential.

66

Chapter 2. Auxiliary Information. Subharmonic Functions

In particular, let m = 2 and μ := n be a zero distribution, i.e., it has unit masses concentrated on a discrete point set {zj : j = 1, 2, . . . }. Then   ∞     z E , p  Π(z, n, p) = log  zj  j=1 where

∞  j=1

 E

z ,p zj



is the canonical Weierstrass product. Theorem 2.9.2.1 (Brelot-Weierstrass) The canonical potential (2.9.2.1) converges uniformly on any bounded domain. It is a subharmonic function with μ as its Riesz measure. Proof. Let |x| < R0 and |y| > R. From the estimate of the primary kernel (Theorem 2.9.1.3) we have       ≤ A(m, p)|x|p+1  G (x, y, m)μ(dy) |y|−p−m+1 μ(dy) p   |y|>R

|y|>R ∞ −p−m+1



= A(m, p)|x|p+1

t

μ(dt).

R

Integrating by part we obtain  ∞ t−p−m+1 μ(dt) = R

μ(R) + (p + m − 1) Rp+m−1



∞

R

μ(t) dt. tp+m

The last integral converges since the genus of μ is p. Hence, both summands tend to zero when R → ∞. Thus      Gp (x, y, m)μ(dy) → 0 sup  |x|
|y|>R

while R0 is fixed and R → ∞, i.e., the canonical potential converges uniformly on any bounded domain. Let us represent the canonical potential for R > R0 in the form   p  G(x, y, Rm )μ(dy) + Dk (x, y)μ(dy) Π(x, μ, p) = |y|
|y|
 +

|y|>R

Gp (x, y, m)μ(dy).

The first summand is a potential, hence a subharmonic function and its Riesz  measure coincide with μ. The other summands are harmonic for |x| < R0 .

2.9. The representation theorem of subharmonic functions in Rm

67

The following proposition estimates the growth of the canonical potential in terms of its masses. Theorem 2.9.2.2 (Estimation of Canonical Potential) The following inequality holds:

 ∞  μ(rτ ) min(1, τ −1 ) μ(r) M (r, Π(•, μ, p)) ≤ A dτ + m−1 (2.9.2.2) rm−2 τ p+m−1 r 0 where A := A(m, p) does not depend on r and μ. Proof. From (2.9.1.10), 

 Π(x, μ, p) ≤ A(m, p)

min Rm

|x|p+1 |x|p , |y|p+m−1 |y|p+m−2

μ(dy).

Set r := |x|, t := |y|. Then we have  p+1  ∞ rp r M (r, Π(•, μ, p)) ≤ A min p+m−1 , p+m−2 μ(dt). t t 0

(2.9.2.3)

The integral on the right side of (2.9.2.3) can be represented in the form  ∞ p+1  r rp r μ(dt) + μ(dt). p+m−2 p+m−1 t t 0 r Integrating every integral by parts we obtain  r  ∞ p+1 μ(r) rp r (p + m − 2) μ(t)dt + (p + m − 1) μ(t)dt + m−1 p+m−1 p+m t r 0 t r  ∞ " r # rp μ(r) ≤ (p + m − 1) min 1, μ(t)dt + m−1 . t tp+m−1 r 0 After the change t = rτ we obtain (2.9.2.2) where the new A(m, p) is equal to A(m, p)(p + m − 1).  Theorem 2.9.2.3 (Brelot-Borel) The order of the canonical potential is equal to the convergence exponent of its mass distribution, i.e., ρ[Π(•, μ, p)] = ρ[μ], if the genus of μ is equal to p. Proof. First assume ρ[μ] < p + 1. Let us choose λ such that ρ[μ] < λ < p + 1. For some constant C that does not depend on t we have μ(t) ≤ Ctλ+m−2 . Actually, μ(t)/tλ+m−2 → 0, because λ > ρ[μ]. Since μ(t) = 0 for small t, this function is bounded and we can take its lower bound as C.

68

Chapter 2. Auxiliary Information. Subharmonic Functions

Now we have f (r, τ ) :=

μ(rτ ) min(1, τ −1 ) ≤ Cτ λ−p−1 min(1, 1/τ ) τ p+m−1

rλ+m−2

(2.9.2.4)

for all τ ∈ (0, ∞). We also have lim f (r, τ ) = 0

r→∞

(2.9.2.5)

because of λ > ρ[μ]. Let us divide (2.9.2.2) by rλ and pass to the upper limit. By Fatou’s lemma (Theorem 2.2.2.3)

 ∞  M (r, Π(•, μ, p)) μ(r) lim sup ≤ A(m, p) lim sup f (r, τ )dτ + lim sup λ+m−1 = 0. rλ r→∞ r→∞ r→∞ r 0 (2.9.2.6) Hence, λ ≥ ρ[Π(•, μ, p)]. (2.9.2.7) Since this holds for any λ > ρ[μ], we have ρ[μ] ≥ ρ[Π(•, μ, p)] under the assumption λ < p[μ] + 1. ¯ Let ρ[μ] = p[μ] + 1. By Theorem 2.8.3.1, Δ[μ] = 0. Hence, μ(t)t−p−m+1 ≤ C and μ(rτ ) min(1, τ −1 ) ≤ C min(1, 1/τ ). f (r, τ ) := (rτ )p+m−1 The function min(1, 1/τ ) is not summable on (0, ∞). Therefore we will act in a slightly different way. From Theorem 2.9.2.2 we have

 1  M (r, Π(•, μ, p)) ≤ A(m, p) lim sup f (r, τ )dτ lim sup rp+1 r→∞ r→∞

0   ∞ μ(t) μ(r) + A(m, p) lim sup dt + + lim sup p+m . tp+m r→∞ r→∞ r r ¯ The first integral is equal to zero because Δ[μ] = 0. The second addend vanishes since the integral converges. Thus we have p + 1 = ρ[μ] ≥ ρ[Π(•, μ, p)]. The reverse inequality holds for any subharmonic function in Rm by the Jensen theorem (Theorem 2.8.3.3).  2.9.3 Let us denote as δSH(ρ) the class of functions u ∈ δSH(Rm ) for which ρT [u] ≤ ρ. Theorem 2.9.3.1 (Brelot-Hadamard) Let u = u1 − u2 ∈ δSH(ρ), and let p1 , p2 be the genuses of the mass distributions μj := μuj , j = 1, 2. Suppose supp[μ1 − μ2 ] ∩ {0} = ∅. Then the following equality holds: u(x) = Π(x, μ1 , p1 ) − Π(x, μ2 , p2 ) + Φq (x) where Φq (x) is a harmonic polynomial of degree q ≤ ρ.

2.9. The representation theorem of subharmonic functions in Rm

69

Proof. The function v(x) := u(x) − Π(x, μ1 , p1 ) + Π(x, μ2 , p2 ) is harmonic by the Brelot-Weierstrass theorem (Theorem 2.9.2.1). We also have the inequality ρT [v] ≤ max(ρT [u], ρT [Π(•, μ1 , p1 )], ρT [Π(•, μ2 , p2 )])

(2.9.3.1)

by Theorem 2.8.2.2 (ρT -properties). The property ρMT1) (Theorem 2.8.2.3) implies ρT [Π(•, μj , pj )] = ρM [Π(•, μj , pj )] := ρ[Π(•, μj , pj )], j = 1, 2. The Brelot-Borel theorem (Theorem 2.9.2.3) implies ρ[Π(•, μj , pj )] = ρ[μj ], j = 1, 2. The Jensen theorem (Theorem 2.8.3.3) implies max(ρ[μ1 ], ρ[μ2 ]) ≤ ρT [u]. From (2.9.3.1) we have ρT [v] ≤ ρT [u] ≤ ρ. Since v is subharmonic, ρT [v] = ρM [v] := ρ[v] by Theorem 2.8.2.3, and ρ[v] ≤ ρ. Therefore M (r, v) =0 lim r→∞ rρ+ for arbitrarily small > 0. By the Liouville theorem (Theorem 2.4.2.3) v(x) is a harmonic polynomial  of degree q ≤ ρ + , and thus v(x) = Φq (x) for q ≤ ρ. For a non-integer ρ the Brelot-Hadamard theorem allows us to connect the growth of functions and masses more tightly than in the Jensen theorem. Theorem 2.9.3.2 (Sharpening of Jensen) Let ρ > 0 and be non-integer, u = u1 − u2 ∈ δSH(Rm ) with ρT [u] = ρ, and let νu = μ1 − μ2 the corresponding charge. Then pJ1) pJ2)

ρ[νu ] = max(ρ[μ1 ], ρ[μ2 ]) = ρ, ¯ u , ρ(r)] ≤ Δ[μ ¯ 1 , ρ(r)] + Δ[μ ¯ 2 , ρ(r)] ≤ A2 σT [u, ρ(r)], A1 σT [u, ρ(r)] ≤ Δ[ν where Aj = Aj (m, ρ) and ρ(r) is an arbitrarily proximate order such that ρ(r) → ρ when r → ∞. For proving this theorem we need

Theorem 2.9.3.3 Let Π(x, μ, p) be a canonical potential with non-integer ρ[μ] := [ρ], and let ρ(r)(→ ρ) be a proximate order. Then ¯ σ[Π(•, μ, p), ρ(r)] ≤ A(m, ρ, p)Δ[μ, ρ(r)].

(2.9.3.2)

70

Chapter 2. Auxiliary Information. Subharmonic Functions

¯ Proof. We can suppose without loss of generality that Δ[μ, ρ(r)] < ∞. By this condition and since μ(t) = 0, 0 < t < c for some c > 0, we have the inequality μ(t)t−ρ(t)−m+2 ≤ C for all t ∈ (0, ∞) and some C > 0 that does not depend on t. Set  ∞ μ(rt) min(1, 1/t) dt. I(r) := ρ(r)+m−2 tp+m−1 c/r r By Theorem 2.9.2.2 we have σ[Π(•, μ, p), ρ(r)] = lim sup r→∞

M (r, Π(•, μ, p) ≤ A(m, p) lim sup I(r). rρ(r) r→∞

(2.9.3.3)

Let us choose r such that μ(r )

sup

(r )ρ(r)+m−2

r>r

¯ ρ(r)] + . ≤ Δ[μ,

For such r we have  ∞ μ(rt) (rt)ρ(rt) min(1, 1/t) dt I(r) = ρ(rt)+m−2 tp+1 rρ(r) c/r (rt)   μ(rt) (rt)ρ(rt) min(1, 1/t) dt ≤ sup ρ(rt)+m−2 tp+1 rρ(r) c/r≤t≤ (rt) c/r  1/  ∞ · · · dt + sup · · · . . . dt + sup · · · ≤t≤1/





≤C c/r

1/≤t≤∞

 ρ(rt)

(rt) rρ(r)

1/

min(1, 1/t) ¯ ρ(r)] + ) dt + (Δ[μ, tp+1





1/

· · · dt + C 

∞

· · · dt.

1/

The function

min(1, 1/t) tp+1  satisfies the conditions of Gol dberg’s theorem (Theorem 2.8.1.5) with p + 1 − ρ < δ < 1 and 0 < γ < p + 1 − ρ. Passing to the limit we have   ¯ ρ(r)] + ) tρ−p dt + (Δ[μ, lim sup I(r) ≤ C f (t) :=

r→∞

0



×



1/

t

ρ−p−1

∞

min(1, 1/t)dt + C



tρ−p−2 dt.

1/

Passing to the limit as → 0 we obtain with the help of (2.9.3.3)  ∞ ¯ ρ(r)] tρ−p−1 min(1, 1/t)dt. σ[Π(•, μ, p), ρ(r)] ≤ A(m, p)Δ[μ, 0



2.9. The representation theorem of subharmonic functions in Rm

71

Proof of Theorem 2.9.3.2. The inequality ρ[νu ] ≤ ρ and the last inequality in pJ2) follow from the Jensen theorem (Theorem 2.8.3.3). Let us prove the reverse inequality and the left side. Since ρ is non-integer, q < ρ in the Brelot-Hadamard theorem (Theorem 2.9.3.1). Hence M (r, Φq ) = o(rρ ) and T (r, u) ≤ T (r, Π(•, μ1 , p)) + T (r, Π(•, μ2 , p)) + o(rρ ). Thus ρT [u] ≤ max(ρ[Π(•, μ1 , p)], ρ[Π(•, μ2 , p]), σT [u, ρ(r)] ≤ max(σT [Π(•, μ1 , p), ρ(r)], σT [Π(•, μ2 , p], ρ(r)]). From Theorem 2.9.3.3 we obtain ρT [u] ≤ max(ρ[μ1 ], ρ[μ2 ]); ¯ 1 , ρ(r)], Δ[μ ¯ 2 , ρ(r)] σT [u, ρ(r)] ≤ A(m, ρ, p) max(Δ[μ ¯ = A(m, ρ, p)Δ[|ν|, ρ(r)]. We can set A1 := A−1 (m, ρ, p) and obtain the left side of pJ2).



2.9.4 Let u ∈ δSH(Rm ) and ρ := ρT [u] be an integer number. We can always represent the function u in the form u(x) = Π(x, ν, ρ) + Φρ (x)

(2.9.4.1)

where Φρ (x) is a harmonic polynomial of degree at most ρ. Actually, such a representation can be obtained from Theorem 2.9.3.1 by addition and subtraction of terms of the form  Dkj (x, y)μj (dy), j = 1, 2 Φkj (x) := Rm

where pj < kj ≤ ρ. All Φkj (x) of such a kind are harmonic polynomials at most ρ. Set  R Π< (x, ν, ρ − 1) := Gρ−1 (x, y, m)ν(dy), |y| (x, ν, ρ) := |y|≥R  δR (x, ν, ρ) := Dρ (x, y)ν(dy).

of degree

(2.9.4.2) (2.9.4.3) (2.9.4.4)

|y|
In particular, for m = 2, δR (z, ν, ρ) :=

1 ρ

 |ζ|


 ρ z ν(dζ). ζ

(2.9.4.4a)

72

Chapter 2. Auxiliary Information. Subharmonic Functions

Let Yρ (x) be the homogeneous polynomial of degree ρ from the polynomial Φρ in (2.9.4.1). Set also δR (x, u, ρ) := δR (x, ν, ρ) + Yρ (x), M (r, δ) := max |δr (ry, u, ρ)|,

(2.9.4.5)

|y|=1

¯ δ [u, ρ] := lim sup M (r, δ)r Δ

−ρ(r)

.

r→∞

The functions δR (x, ν, ρ) are homogeneous polynomials that are determined completely by their values on the unit sphere. Thus, by the Harnack theorem (Theorem 2.4.1.7) we have ¯ δ [u, ρ(r)] < ∞ if and only if the family δR (x, u, ρ)Rρ−ρ(R) , R > Theorem 2.9.4.1 Δ 0 is precompact in D (Rm ). Let ρ be an integer number and ρ(r) → ρ be a p.o. Set ¯ u |, ρ(r)]. ¯ δ [u, ρ(r)], Δ[|ν Ω[u, ρ(r)] := max(Δ Theorem 2.9.4.2 (Brelot-Lindel¨ of ) The following holds: A1 Ω[u, ρ(r)] ≤ σT [u, ρ(r)] ≤ A2 Ω[u, ρ(r)], where Aj := Aj (m, ρ). R For proving this theorem we will first study the function ΠR < and Π> . Set

T (r, λ, >) := T (r, Πλr > (•, ν, ρ)), T (r, λ, <) := T (r, Πλr < (•, ν, ρ − 1)). Theorem 2.9.4.3 (Estimate of T (•, >, T (•, <)) The following holds:  ∞ |ν|(rt) min(1, t−1 ) |ν|(r) T (r, λ, >) ≤ A dt + , rm−2 tρ+m−1 rm−1 λ % $ λ |ν|(rt) min(1, t−1 ) dt T (r, λ, <) ≤ A rm−2 tρ+m−2 0   |ν|(rλ) ∞ min(1, t−1 ) |ν|(r) dt + m−1 , +A rm−2 λ tρ+m−2 r

(2.9.4.6)

(2.9.4.7)

where A := A(m, ρ). Proof. Let ν = μ1 − μ2 . Then |ν| = μ1 + μ2 . We have R R ΠR < (x, ν, ρ − 1) = Π< (x, μ1 , ρ − 1) − Π< (x, μ2 , ρ − 1).

(2.9.4.8)

Since ΠR < (0, μ2 , ρ − 1) = 0, we have (see t3),t4), Theorem 2.8.2.1) R R T (r, ΠR < (•, ν, ρ − 1)) ≤ T (r, Π< (•, μ1 , ρ − 1)) + T (r, Π< (•, μ2 , ρ − 1)). (2.9.4.9)

2.9. The representation theorem of subharmonic functions in Rm

73

Set R Π1 := ΠR < (•, μ1 , ρ − 1), Π2 := Π< (•, μ2 , ρ − 1).

Let us estimate, for example, T (r, Π1 ). The masses of the canonical potential Π1 are concentrated in KR . Applying Theorem 2.9.2.2 (Estimation of Canonical Potential) for p = ρ − 1 we obtain T (r, Π1 ) ≤ M (r, Π1 )  Rr  μ1 (R) ∞ min(1, t−1 ) μ1 (r) μ1 (rt) min(1, t−1 ) ≤A dt + A m−2 dt + m−1 . m−2 ρ+m−2 ρ+m−2 R r t r t r 0 r Set R := rλ. Then we obtain the inequality (2.9.4.7) for ν := μ1 . Analogously one can do the same for ν := μ2 . The inequality (2.9.4.9) allows us to pass to the limit in (2.9.4.7) in the general case. Set Π1 := ΠR > (•, μ1 , ρ). Applying (2.9.2.2) for p = ρ we obtain  T (r, Π1 ) ≤ M (r, Π1 ) ≤

∞ R r

μ1 (rt) min(1, t−1 ) μ1 (r) dt + m−1 . rm−2 tρ+m−1 r 

In the same way we obtain (2.9.4.6). Set T (r, Πr> (•, ν, ρ)) , rρ(r) r→∞ T (r, Πr< (•, ν, ρ)) . σ[Π< , ρ(r)] := lim sup rρ(r) r→∞ σ[Π> , ρ(r)] := lim sup

Theorem 2.9.4.4 Let ν := μ1 − μ2 ∈ δM(ρ) and ρ an integer number. Then for any p.o. ρ(r) → ρ, ¯ ρ(r)] max(σ[Π> , ρ(r)], σ[Π< , ρ(r)]) ≤ AΔ[|ν|, where A := A(m, ρ). Proof. From (2.9.4.6) we have  T (r, Πr> (•, ν, ρ)) = T (r, 1, >) ≤ A 1

∞

|ν|(rt) 1 |ν|(r) dt + m−1 . rm−2 tρ+m r

Now we repeat the reasoning of Theorem 2.9.3.3 for μ := |ν| and p := ρ. We will obtain  ∞ ¯ σ[Π> , ρ(r)] ≤ AΔ[|ν|, ρ(r)] t−2 dt. 1

74

Chapter 2. Auxiliary Information. Subharmonic Functions

For the other case we have from (2.9.4.7), T (r, Πr< (•, ν, ρ − 1)) = T (r, 1, <)  ∞  1 |ν|(rt) 1 |ν|(r) −ρ−m+1 −1 dt + A t dt + r ≤A . m−2 tρ+m−1 rm−2 0 r 1 We divide this inequality by rρ(r) and pass to the upper limit while r → ∞. The first summand of the right side gives  1 ¯ AΔ[|ν|, ρ(r)] dt 0

by the reasoning of Theorem 2.9.3.3. The second one can be computed directly, yielding  ∞ ¯ t−ρ−m+1 dt. AΔ[|ν|, ρ(r)] 1

Combining all these inequalities we obtain the assertion of the theorem.



Proof of Theorem 2.9.4.2. Let us represent u(x) in the form u(ry) = Πr< (ry, νu , ρ − 1) + Πr> (ry, νu , ρ) + δr (ry, u, ρ) + o(rρ−1 )

(2.9.4.10)

where |y| = 1. Then we have T (r, u) ≤ T (r, Πr< (•, νu , ρ − 1)) + T (r, Πr> (•, νu , ρ)) + M (r, δ) + o(rρ−1 ). Let us divide this by rρ(r) and pass to the upper limit. By Theorem 2.9.4.4 we obtain ¯ ¯ δ [u, ρ(r)]) = A2 Ω[u, ρ(r)] ρ(r)], Δ σT [u, ρ(r)] ≤ A max(Δ[|ν|, where A2 = A(m, ρ). Let us write (2.9.4.11) in the form δr (ry, u, ρ) = u(ry) − Πr< (ry, νu , ρ − 1) − Πr> (ry, νu , ρ) + o(rρ−1 ). We obtain T (r, δr (•, u, ρ)) ≤ T (r, u) + T (r, Πr< (•, νu , ρ − 1)) + T (r, Πr> (•, νu , ρ) + o(rρ−1 ). Since δR (•, u, ρ) is harmonic and homogeneous, we have by (2.8.2.5) M (r, δR ) ≤ 2m−1 T (2r, δR ) = 2m−1+ρ T (r, δR ). Therefore we obtain the inequality ¯ δ [u, ρ(r)] ≤ σT [u, ρ(r)] + 2AΔ[|ν|, ¯ Δ ρ(r)]. By the Jensen theorem (Theorem 2.8.3.3) we have Ω[u, ρ(r)] ≤ A−1 1 σT [u, ρ(r)] for some A1 = A1 (m, ρ).



Chapter 3

Asymptotic Behavior of Subharmonic Functions of Finite Order 3.1 Limit sets 3.1.1 Let {Vt : t ∈ (0, ∞)} be a family of rotations of Rm that form a oneparametric group, i.e., (3.1.1.0) Vt1 Vt2 = Vt1 t2 , V1 = I, where I is the identity map. The family of linear transformations Pt := tVt

(3.1.1.1)

is also a one-parametric group. In particular, for m = 2 the general form of the rotations is Vt z = z exp(iγ log t), where γ is real. The orbit {Pt z : t ∈ (0, ∞)} of every point z = 0 is a logarithmic spiral if γ = 0 and a ray if γ = 0. For m ≥ 3 and Vt ≡ I, t ∈ (0, ∞) the orbit of every point x = 0 is a ray from the origin. For other V• it is a spiral connecting the origin to infinity. It is clear that only one orbit {Pt x : t ∈ (0, ∞)} passes through every x = 0. The behavior of every point y(t) := Pt x is completely determined by a system of

76

Chapter 3. Asymptotic Behavior of Subharmonic Functions of Finite Order

differential equations with constant coefficients: d d y = (I + V  )y, V  := Vt |t=1 dt dt with the initial condition of y(1) = x. 3.1.2 Let u ∈ SH(ρ) and σM [u, ρ(r)] < ∞ for some p.o. ρ(r) → ρ. We will write u ∈ SH(Rm , ρ, ρ(r)) or shorter, u ∈ SH(ρ(r)). For u ∈ SH(ρ(r)) set ut (x) := u(Pt x)t−ρ(t) .

(3.1.2.1)

We will denote this transformation as (•)t . Theorem 3.1.2.1 (Existence of Limit Set) The following holds: els1)

ut ∈ SH(ρ(r)) for any t ∈ (0, ∞),

els2)

the family {ut } is precompact at infinity.

I.e., for any sequence tk → ∞ there exists a subsequence tkj → ∞ and a function v ∈ SH(Rm ) such that utkj → v in D (Rm ) (see Section 2.7.1). Proof. The functions ut are subharmonic by sh1) and sh5), Theorem 2.6.1.1. (Elementary Properties), and M (r, ut ) = M (rt, u)t−ρ(t) . Now we have σM [ut , ρ(r)] = t−ρ(t) lim sup r→∞

because

M (rt, u) (rt)ρ(rt) · lim = σM [u, ρ(r)]tρ−ρ(t) , ρ(rt) r→∞ rρ(r) (rt)

(rt)ρ(rt) L(rt) = tρ = tρ lim r→∞ rρ(r) r→∞ L(r) lim

(3.1.2.2)

(see, ppo3), Theorem 2.8.1.3 (Properties of P.O)). Therefore els1) is proved. Let us check the conditions of Theorem 2.7.1.1 (Compactness in D ). We have lim sup M (r, ut ) = lim sup t→∞

t→∞

M (rt, u) (rt)ρ(rt) · lim = σM [u, ρ(r)]rρ . (rt)ρ(rt) t→∞ tρ(t)

(3.1.2.3)

Thus, the family is bounded from above on every compact set and lim ut (0) = lim u(0)t−ρ(t) = 0.

t→∞

t→∞

Therefore ut (0) are bounded from below for large t.



3.1. Limit sets

77

We will call the set of all functions v from Theorem 3.1.2.1 the limit set of the function u(x) with respect to V• and denote it by Fr[u, ρ(r), V• , Rm ] or shortly Fr[u]. The limit set does not depend on values of the subharmonic function on a bounded set, hence, it is a characteristic of asymptotic behavior. Set U [ρ, σ] := {v ∈ SH(Rm ) : M (r, v) ≤ σrρ , r ∈ [0, ∞); v(0) = 0},  U [ρ, σ] (3.1.2.4) U [ρ] := σ>0

and

v[t] (x) := t−ρ v(Pt x), t ∈ (0, ∞).

(3.1.2.4a)

Let us emphasize that the transformation (•)[t] coincides with (•)t from (3.1.2.1) for ρ(r) ≡ ρ and satisfies the condition (•)[tτ ] = ((•)[t] )[τ ]

(3.1.2.4b)

Theorem 3.1.2.2 (Properties of Fr) The following holds: fr1) Fr[u] is a connected compact set; fr2) Fr[u] ⊂ U [ρ, σ], for σ ≥ σM [u]; fr3) (Fr[u])[t] = Fr[u], t ∈ (0, ∞).

I.e., v ∈ Fr[u] implies v[t] ∈ Fr[u];

fr4) if ρ1 (r) and ρ(r) are equivalent (see (2.8.1.5)), then Fr[u, ρ1 (r), •] = Fr[u, ρ(r), •]. We need the following assertion. Theorem 3.1.2.3 (Continuity ut ) The functions ut , v[t] : (0, ∞) × D (Rm ) → D (Rm ) are continuous in the natural topology. Proof. For any ψ ∈ D(Rm ) consider   ut , ψ := ut (x)ψ(x)dx = u(y)ψ(y/t)tm−ρ(t) dy := u, ψ(•, t), where ψ(y, t) := ψ(y/t)tm−ρ(t) . The function ψ(•, t) is continuous in t in D(Rm ). By Theorem 2.3.4.6 (Continuity •, •) u, ψ(•, t) is continuous in (u, t). 

78

Chapter 3. Asymptotic Behavior of Subharmonic Functions of Finite Order

Proof of Theorem 3.1.2.2. Let us denote as clos{•} the closure in D -topology. The set FN := clos{ut : t ≥ N } ⊃ Fr[u] is compact in D -topology. Indeed, let tj → t and t < ∞; then utj → ut because of Theorem 3.1.2.3. If tj → ∞ and utj → v, then v ∈ Fr[u] by its definition, hence, v ∈ FN . Since Fr[u] = ∩∞ N =1 FN , it is compact. Let us prove the connectedness. Suppose Fr[u] is not connected. Then it can be written as a union of two disjoint nonempty closed sets F 1 and F 2 . Let V 1 , V 2 be disjoint open neighborhoods of F 1 , F 2 respectively in D (Rm ). Since F 1 , F 2 are nonempty there exist sequences {sj }, {tj } such that sj < tj , sj → ∞, usj ∈ V 1 , utj ∈ V 2 . Since the mapping ut : (0, ∞) → D (Rm ) is continuous, by Theorem 3.1.2.3 its image is connected. Thus there exists a sequence {pj } with sj < pj < tj such that upj ∈ / V 1 ∪ V 2 . This sequence has a subsequence that converges to a function v ∈ Fr[u] and v ∈ / F 1 ∪ F 2 . This is a contradiction. Hence, Fr[u] is connected and fr1) is proved. Set ψ(r) := lim sup M (r, ut ). r→∞

This function is convex with respect to −r2−m for m ≥ 3 and with respect to log r for m = 2 and hence continuous. Indeed, M (|x|, ut ) are subharmonic (see Theorem 2.6.5.2 (Convexity M (•, u) and M(r, u)). By Theorem 2.7.3.3 (H.Cartan +) the function ψ ∗ (|x|) is subharmonic and ψ(|x|) = ψ ∗ (|x|) quasi-everywhere. However, if ψ(|x|) < ψ ∗ (|x|) at some point, the same inequality holds on a sphere which has a positive capacity (see Example 2.5.2.2). Hence, ψ(|x|) = ψ ∗ (|x|) everywhere, and ψ(|x|) is subharmonic. Thus ψ(r) is convex with respect to −r2−m for m ≥ 3 and with respect to log r for m = 2 by Theorem 2.6.3.2 (Subharmonicity and Convexity). One can also see that for u ∈ SH(Rm ), M (r, u ) ≤ M (r + , u), where (•) is defined by (2.6.2.3). Let v ∈ Fr[u] and utj → v in D (Rm ). By property reg3), Theorem 2.3.4.5 (utj ) → v uniformly on any compact set. Thus v (x) = lim (utj ) ≤ lim sup M (|x|, (ut ) ) j→∞

t→∞)

≤ lim sup M (|x| + , ut ) = ψ(|x| + ).

(3.1.2.5)

t→∞)

If ↓ 0, then v ↓ v by Theorem 2.6.2.3 and ψ(r + ) → ψ(r) because of continuity. Passing to the limit in (3.1.2.5) and using (3.1.2.3) we obtain v(x) ≤ σM [u, ρ(r)]|x|ρ .

(3.1.2.6)

3.1. Limit sets

79

Since u(0) ≤ u (0) we have u(0)t−ρ(t) ≤ (ut ) (0). Let us pass to the limit as t := tj → ∞. We obtain v (0) ≥ 0. Passing to the limit as ↓ 0 we have v(0) ≥ 0.

(3.1.2.7)

The inequalities (3.1.2.6) and (3.1.2.7) imply fr2). One can check the equality (ut )[τ ] = utτ ·

(tτ )ρ(tτ ) . tρ(t) τ ρ

(3.1.2.8)

By using properties of p.o. we have (tτ )ρ(tτ ) =1 t→∞ tρ(t) τ ρ lim

(compare (3.1.2.2)). Let v ∈ Fr[u] and utj → v. Set t := tj , τ := t in (3.1.2.8) and pass to the limit. Then v[t] = D − lim utj t . j→∞

Thus v[t] ∈ Fr[u]. The property f3) is proved. Let us prove f4).We have u(Pt x) u(Pt x) u(Pt x) = ρ(t) × e(ρ1 (t)−ρ(t)) log t = ρ(t) × (1 + o(1)) ρ (t) 1 t t t as t → ∞ because of (2.8.1.5). This implies f4). 

Exercise 3.1.2.1 Check this in detail.

We can consider the limit sets as a mapping u → Fr[u]. The following theorem describes some properties of this mapping. Set  U [ρ, σ] (3.1.2.9) U [ρ] := σ>0

where U [ρ, σ] is defined by (3.1.2.4). Let X, Y be subsets of a cone (i.e., a subset of a linear space that is closed with respect to sum and multiplication by a positive number). The set U [ρ] is such a cone. Set X + Y := {z = x + y : x ∈ X, y ∈ Y }; λX := {z = λx : x ∈ X}. Theorem 3.1.2.4 (Properties of u → Fr[u]) The following holds: fru1)

Fr[u1 + u2 ] ⊂ Fr[u1 ] + Fr[u2 ],

fru2)

Fr[λu] = λFr[u].

(3.1.2.10)

80

Chapter 3. Asymptotic Behavior of Subharmonic Functions of Finite Order

Proof. Let v ∈ Fr[u1 + u2 ]. Then there exists tj → ∞ such that (u1 + u2 )tj → v in D . We can find a subsequence tjk such that (u1 )tjk → v1 and (u2 )tjk → v2 . Then v = v1 + v2 . The property fru1) has been proved. The property fru2) is proved analogically.  3.1.3 We will write μ ∈ M(Rm , ρ(r)) or shortly, μ ∈ M(ρ(r)) if μ ∈ M(Rm ) (see ¯ 2.8.3) and Δ[μ, ρ(r)] < ∞ (see 2.8.3.2). Let us define a distribution μt for μ ∈ M(ρ(r)) by  −ρ(t)−m+2 (3.1.3.1) φ(Pt−1 x)μ(dx) μt , φ := t for φ ∈ D(Rm ). It is positive. Hence, it defines uniquely a measure μt . Theorem 3.1.3.1 (Explicit form of μt ) For any E ∈ σ(Rm ) the following holds: μt (E) = t−ρ(t)−m+2 μ(Pt E).

(3.1.3.2)

Proof. It is enough to prove the assertion for some dense ring (see Theorem 2.2.3.5), for example, for all compact sets. Let χK be a characteristic function of a compact set K and let φ ↓ χK be a monotonically converging sequence of functions that belong to D(Rm ) (see Theorems 2.1.2.1, 2.1.2.9 and 2.3.4.4). Then   −ρ(t)−m+2 φ (x)μt (dx) = t φ (Pt−1 x)μ(dx). Since φ (Pt−1 x) ↓ χPt K (x),   −ρ(t)−m+2 χPt K (x)μ(dx) = t−ρ(t)−m+2 μ(Pt K). μt (K) = χK (x)μt (dx) = t



Theorem 3.1.3.2 (Existence of μ-Limit Set) The following holds: mels1) μt ∈ M(ρ(r)) for any t ∈ (0, ∞); mels2) the family {μt } is precompact in infinity. I.e., for any sequence tk → ∞ there exists a subsequence tkj → ∞ and a measure ν ∈ M(Rm ) such that μtkj → ν in D (Rm ) (see Section 2.7.1). Proof. We have

μt (r) = μ(rt)t−ρ(t)−m+2 .

Thus lim sup r→∞

μt (r) rρ(r)+m−2

= lim sup r→∞

=t Therefore mels1) holds.

μ(rt)

(rt)ρ(rt)

(rt)ρ(rt)+m−2 tρ(t)+m−2 rρ(r)

ρ−ρ(t)−(m−2)

¯ Δ[μ, ρ(r)].

(3.1.3.3)

3.1. Limit sets

81

We also have ¯ ρ(r)]rρ+m−2 . lim sup μt (r) = Δ[μ, t→∞

Thus μt satisfies the assumption of the Helly theorem (Theorem 2.2.3.2). Using also Theorem 2.3.4.4 we obtain mels2).  We will call the set of all measures ν from Theorem 3.1.2.1 the limit set of the mass distribution μ with respect to V• and denote it Fr[μ, ρ(r), V• , Rm ] or shortly, Fr[μ]. Set (3.1.3.4) M[ρ, Δ] := {ν : ν(r) ≤ Δrρ+m−2 , ∀r > 0}.  M[ρ, Δ], M[ρ] := Δ>0

and

ν[t] (E) := t−ρ−m+2 ν(Pt E)

(3.1.3.5)

for E ∈ σ(R ). m

Theorem 3.1.3.3 (Properties of Fr[μ]) The following holds: frm1) Fr[μ] is connected and compact; ¯ ρ(r)]; frm2) Fr[μ] ⊂ M[Δ, ρ], for Δ ≥ Δ[μ, frm3) (Fr[μ])[t] = Fr[μ], t ∈ (0, ∞). Proof. We will only prove frm2) because frm1) and frm3) are proved word by word as in Theorem 3.1.2.2. Suppose tn → ∞ and μtn → ν ∈ Fr[μ]. Let us choose r > r such that the open ball Kr is squarable with respect to ν. It is possible because of Theorem 2.2.3.3, sqr2). By Theorems 2.2.3.7 and 2.3.4.4, μtn (r ) → ν(r ). Thus (compare with (3.1.2.3)) ¯ ρ(r)](r )ρ+m−2 . ν(r ) = lim μtn (r ) ≤ lim sup μt (r ) = Δ[μ, tn →∞

t→∞



Choosing r ↓ r we obtain

lim ν(r ) = ν(r)

r  →r

because (2.2.3.3). Thus frm2) holds. The following assertion is a “copy” of Theorem 3.1.2.4. Theorem 3.1.3.4 (Properties of μ → Fr[μ]) The following holds: frmu1) Fr[μ1 + μ2 ] ⊂ Fr[μ1 ] + Fr[μ2 ], frmu2) Fr[λμ] = λFr[μ]. The proof is also a “copy” and we omit it. Exercise 3.1.3.1 Prove Theorem 3.1.3.4



82

Chapter 3. Asymptotic Behavior of Subharmonic Functions of Finite Order

3.1.4 We are going to study the class U [ρ] and obtain for it “non-asymptotic” analogies of Theorem 2.8.3.3 (Jensen), 2.9.2.3 (Brelot-Borel), 2.9.3.1 (Brelot-Hadamard) Theorem 3.1.4.1 (*Jensen) Let v ∈ U [ρ]. Then its Riesz measure νv ∈ M[ρ]. Proof. As in Theorem 2.8.3.2 we have an inequality νv (r) ≤ A(m)N (2r, νv ). rm−2

(3.1.4.1)

Since v(0) = 0 we have (Theorem 2.6.5.1. (Jensen-Privalov)) N (2r, νv ) = M(2r, v) ≤ M (2r, v) ≤ 2ρ σrρ .

(3.1.4.2)

Substituting (3.1.4.2) in (3.1.4.1) we obtain νv ∈ M[ρ, Δ] for some Δ. Thus νv ∈ M[ρ].  Let ρ be non-integer and ν ∈ M[ρ]. Consider the canonical potential Π(x,ν,p) where p := [ρ] (see (2.9.2.1)). Let us emphasize that the support of ν may contain the origin but ν(0) = 0, i.e., there is no concentrated mass in the origin. Thus we must also check its convergence in the origin. Theorem 3.1.4.2 (*Brelot-Borel) Let ρ be non-integer and let ν ∈ M[ρ]. Then Π(x, ν, p) converges and belongs to U [ρ]. Proof. Using (2.9.1.9) we have    2|x|   ν(dt)   Gp (x, y, m)ν(dy) ≤ A(m, p)|x|p .  p+m−2   |y|<2|x| t 0

(3.1.4.3)

Let us estimate the integral in (3.1.4.3). Integrating by parts we obtain  2|x|  2|x| ν(dt) ν(t)dt ν(t) 2|x| I< (x) := = | + (p + m − 2) . 0 p+m−2 p+m−2 t t tp+m−1 0 0 Since ν ∈ M[ρ, Δ] for some Δ, I< (x) ≤ A(m, ρ, p)Δ|x|ρ−p . Substituting this in (3.1.4.3) we obtain       Gp (x, y, m)ν(dy) ≤ A(m, ρ, p)Δ|x|ρ .    |y|<2|x| Analogously, using (2.9.1.8) we obtain       Gp (x, y, m)dν(dy) ≤ A(m, ρ, p)Δ|x|ρ .    |x|<2|y|

(3.1.4.4)

(3.1.4.5)

3.1. Limit sets

83

In particular, these estimates show that Π(x, ν, p) exists. Now using (2.9.1.10) we have also   2|x|  |x|p+1 |x|p Gp (x, y, m)ν(dy) ≤ A(m, p) ν(dt) min p+m−1 , p+m−2 . |x| |x| t t 2 ≤|y|≤2|x| 2 The latter integral can also be easily estimated by ΔA(m, p, ρ)|x|ρ . Thus we have  Gp (x, y, m)ν(dy) ≤ A(m, p, ρ)|x|ρ . |x| 2 ≤|y|≤2|x|

Therefore by (3.1.4.5) and (3.1.4.4) we obtain M (r, Π) ≤ σrρ for some σ. Since Gp (0,y,m) = 0 for all y = 0 and the integral converges, Π(0,ν,p) = 0.



We will need an assertion that looks like the Liouville theorem (Theorem 2.4.2.3). Theorem 3.1.4.3 (*Liouville) Let H be a harmonic function in Rm and H ∈ U [ρ]. Then H ≡ 0 if ρ is non-integer and H is a homogeneous polynomial of degree p if ρ = p is integer. In particular, for m = 2 we have H(reiφ ) = rp (ceipφ ). Proof. Like in the proof of the Liouville theorem we obtain the inequality (2.4.2.9) and |ck | ≤ AR−k max H(x) ≤ AσRρ−k |x|=R

for some σ > 0. If k > ρ, we will pass to the limit when R → ∞ and obtain ck = 0. If k < ρ, we will do that when R → 0 and obtain ck = 0.  The following theorem can be considered as an analogy of the Brelot-Hadamard theorem (Theorem 2.9.3.1): Theorem 3.1.4.4 (*Hadamard) Let ρ be non-integer and v ∈ U [ρ]. Then v(x) = Π(x, νv , p)

(3.1.4.6)

for p = [ρ]. Proof. Consider the function H(x) := v(x) − Π(x, νv , p). It is harmonic.We also have by(2.8.2.5) M (r, H) ≤ A(m)T (r, H) ≤ A(m)[T (r, v) + T (r, Π)] ≤ σrρ for some σ. Hence, H(x) ≡ 0 by Theorem 3.1.4.3.



84

Chapter 3. Asymptotic Behavior of Subharmonic Functions of Finite Order

Let us consider the case of integer ρ. Let ν ∈ M[ρ] for an integer ρ = p. Set  Gp−1 (x, y, m)ν(dy), (3.1.4.7) Π< (x, ν, ρ) := |y|<1  Π> (x, ν, ρ) := Gp (x, y, m)ν(dy). (3.1.4.8) |y|≥1

Both potentials converge and belong to U [ρ]. Theorem 3.1.4.5 (**Hadamard) Let ρ be integer and let v ∈ U [ρ]. Then v = Hρ (x) + Π< (x, ν, ρ) + Π> (x, ν, ρ),

(3.1.4.9)

where Hρ is a homogeneous harmonic polynomial of degree ρ. The proof is exactly the same as in the *Hadamard theorem, but we use the second case of Theorem 3.1.4.3. We also note that the polynomial may be equal to zero identically. Exercise 3.1.4.1 Check this in detail. Let as check that ν from (3.1.4.9) has the following property that is analogous to Theorem 2.9.4.2. Theorem 3.1.4.6 (*Lindel¨ of ) Let ρ be integer and let v ∈ U [ρ]. Then  Dρ (x, y)μ(dy) = Hρ (x). lim →0 ≤|y|<1

Proof. Consider the function ⎧  ν(dy) ⎪ |x−y|m−2 f or m > 2; ⎨ |y|<  v∗ (x) := v(x) + ⎪ log |x − y|ν(dy), f or m = 2. ⎩−

(3.1.4.10)

(3.1.4.11)

|y|<

It is subharmonic with supp ν∩{0} = ∅. We represent this function as in (2.9.4.10) in the form ∗ (x, v∗ ) + δ1 (x, v∗ , ρ). v∗ (x) = Π1< (x, ν∗ , ρ) + Π1> (x, ν∗ , ρ) + Pρ−1

In this representation we can pass to the limit as → 0 in the left side and in all the summands except perhaps the last two from the right side. Exercise 3.1.4.2 Check this, using that all the integrals converge for ν ∈ M(ρ, Δ) and showing that the integral in (3.1.4.11) tends to zero. The last two summands form a harmonic polynomial, the limit of which is also a harmonic polynomial. Comparing the limit with the representation (3.1.4.9), ∗ we obtain that Pρ−1 (•, v∗ ) tends to zero and δ1 (x, v∗ , ρ) tends to Hρ (x). 

3.1. Limit sets

85

Theorem 3.1.4.7 (**Liouville) If v ∈ U [ρ] satisfies inequality v(x) ≤ 0 for z ∈ Rm , then v(x) ≡ 0. Otherwise it contradicts subharmonicity in 0. 3.1.5 Let us study the connection between Fr[u] and Fr[μu ]. Note the following properties of the transformations (•)t and (•)[t] . Theorem 3.1.5.0 (Connection between ut and μt ) One has (μu )t = μut ; (μv )[t] = μv[t] .

(3.1.5.0)

Proof. By the F. Riesz theorem (Theorem 2.6.4.3) and Theorem 2.5.1.1, GPo3) we have for any ψ ∈ D(Rm ), μu , ψ = θm Δu, ψ = θm u, Δψ. Using the definition (3.1.3.1), we obtain (μu )t , ψ = (μu )t , ψ((Pt )−1 •)t−ρ(t)−m+2 . Thus (μu )t , ψ = θm u, Δ[ψ((Pt )−1 •)]t−ρ(t)−m+2 . Since the Laplace operator is invariant with respect to Vt for any t we have Δ[ψ((Pt )−1 •)] = t−2 [Δψ]((Pt )−1 •). Thus we obtain (μu )t , ψ = θm t−ρ(t)−m u, [Δψ]((Pt )−1 •) = θm u(Pt •)t−ρ(t) , Δψ = θm ut , Δψ = μut , ψ.



Exercise 3.1.5.1 Do this for (•)[t] . We begin from the case of a non-integer ρ. Theorem 3.1.5.1 (Connection between Fr’s for non-integer ρ) Let u ∈ U (ρ(r)) and μu be its Riesz measure. Then Fr[μu ] = {νv : v ∈ Fr[u]}, Fr[u] = {Π(•, ν, p) : ν ∈ Fr[μu ]}.

(3.1.5.1) (3.1.5.2)

Proof. Let ν ∈ Fr[μu ]. There exists tn → ∞ such that (μu )tn → ν in D . We can find a subsequence tn such that utn → v ∈ Fr[u]. Thus (μu )tn → νv and therefore

86

Chapter 3. Asymptotic Behavior of Subharmonic Functions of Finite Order

ν = νv . Hence, Fr[μu ] ⊂ {νv : v ∈ Fr[u]}. Analogously we can prove that every νv ∈ Fr[μu ] and hence (3.1.5.1) holds. Let ν ∈ Fr[μu ]. We find a sequence tn → ∞ such that (μu )tn → ν in D . We find a subsequence tn such that utn → v ∈ Fr[u] and νv = ν. By the *Hadamard theorem (Theorem 3.1.4.4) v = Π(•, ν, p). Hence, {Π(•, ν, p) : ν ∈ Fr[μu ]} ⊂ Fr[u]. And vice versa, since Fr[u] ⊂ U [ρ] (Theorem 3.1.2.2, fr2)), every v ∈ Fr[u] is represented as Π(•, νv , p) and νv ∈ Fr[μ] by (3.1.5.1).  Let ρ be integer and u ∈ U (ρ(r)). Let us consider the precompact family of homogeneous polynomials δt (x, u, ρ)tρ−ρ(t) from Theorem 2.9.4.1. For every tn → ρ−ρ(tn ) ∞ we can find a subsequence tn such that the pair (δtn (•, u, ρ)tn , (μu )tn ) tends to a pair (Hν , ν) where Hν is a homogeneous harmonic polynomial of degree p. We denote the set of all such pairs as (H, Fr)[u]. Every v ∈ U [ρ] can be represented in the form (3.1.4.7). Thus for every v the polynomial H v := Hp is determined. Theorem 3.1.5.2 (Connection between Fr’s for integer ρ) Let u ∈ U (ρ(r)). Then (H, Fr)[u] = {(H v , νv ) : v ∈ Fr[u]}, Fr[u] = {v := Hν + Π< (•, ν, ρ) + Π> (•, ν, ρ) : (Hν , ν) ∈ (H, Fr)[u]}.

(3.1.5.3) (3.1.5.4)

The proof is clear. 3.1.6 Up to now we supposed that the family of rotations V• was fixed. Now we take in consideration that it can vary and use the notation Fr[u, V• ]. Theorem 3.1.6.1 (Dependence of Fr on V• ) Let Fr[u, V• ] and Fr[u, W• ] be limit sets of u with respect to rotation families V• and W• accordingly. Then for any v ∈ Fr[u, V• ] there exist a rotation V v and wv ∈ Fr[u, W• ] such that v(x) = wv (V v x) for all x ∈ Rm . Proof. Let v ∈ Fr[u, V• ] and let tn → ∞ be a sequence such that tn−ρ(tn ) u(tn Vtn •) → v. Since the family Vt is obviously precompact there exists a subsequence (for which we keep the same notation), and a rotation V v such that Wt−1 Vtn → V v and n −ρ(tn ) u(tn Wtn •) → w. w ∈ Fr[u, W• ] such that tn Now we have v(•) = D − lim tn−ρ(tn ) u(tn Vtn •) Vtn •) = D − lim tn−ρ(tn ) u(tn Wtn Wt−1 n = w(V v •).



3.2. Indicators

87

3.2 Indicators 3.2.1 Let u ∈ SH(ρ(r)) and let Fr[u] be the limit set. Set h(x, u) := sup{v(x) : v ∈ Fr[u]},

(3.2.1.1)

h(x, u) := inf{v(x) : v ∈ Fr[u]}.

(3.2.1.2)

These functionals reflect the asymptotic behavior of u along curves of the form lx0 := {x = tVt x0 : t ∈ (0, ∞)}

(3.2.1.3)

and are called indicator of growth of u and lower indicator respectively. Of course, the indicators depend on ρ(r) and Vt , but we will only note that if necessary. Theorem 3.2.1.1 (Properties of Indicators) The following holds: h1) h is upper semicontinuous, h is subharmonic; h2) they are semiadditive and positively homogeneous, i.e., h(x, u1 + u2 ) ≤ h(x, u1 ) + h(x, u2 );

(3.2.1.4)

h(x, u1 + u2 ) ≥ h(x, u1 ) + h(x, u2 ); h, h(x, Cu) = Ch, h(x, u), C ≥ 0;

(3.2.1.5) (3.2.1.6)

h, h[t] (x, u) = h, h(x, u).

(3.2.1.7)

h3) invariance:

Proof. Semicontinuity of h follows from Theorem 2.1.2.8 (Commutativity of inf and M(.). Semicontinuity and subharmonicity of h follow from Theorem 2.7.3.4 (Sigurdsson’s Lemma). The properties h2) follow from properties of infimum and supremum. The invariance follows from invariance of Fr[u] (Theorem 3.1.2.2, fr3)).  Set −1 (x) x0 (x) := P|x|

(3.2.1.8)

where Pt is defined by (3.1.1.1). This is an intersection of the orbit of Pt that passes through a point x with the unit sphere. If Vt ≡ I, (3.2.1.9) x0 (x) = x/|x| := x0 .

88

Chapter 3. Asymptotic Behavior of Subharmonic Functions of Finite Order

Theorem 3.2.1.2 (Homogeneity h, h) One has h, h(x, •) = |x|ρ h, h(x0 (x), •)

(3.2.1.10)

Thus the indicators are determined uniquely by their values on the unit sphere, i.e., they are “functions of direction”. In particular, they are homogeneous for Vt ≡ I: h, h(x, •) = |x|ρ h, h(x0 , •). (3.2.1.11) The proof of (3.2.1.10) follows from h3), Theorem 3.2.1.1 if we set t := |x|; x := Pt−1 x. 3.2.2 In this item we will suppose that Vt ≡ I and study the indicator. Let Δx0 be as defined in Section 2.4.1. Its coefficients depend on a choice of the spherical coordinate system. However, one has Theorem 3.2.2.1 Let ψ(y) have continuous second derivatives on the unit sphere S1 . Then the differential form Δx0 ψ(y)dy is invariant with respect to the choice of spherical coordinate system. Proof. Let φ(x) be a smooth function in Rm . Then Δφ(x)dx is invariant with respect to the choice of an orthogonal system because Δ (the Laplace operator) and an element of volume are invariant. Set φ(x) = ψ(y), where y := x0 = x/|x|. Then Δφdx = Δx0 ψ(y)dy rm−3 dr. Since r is invariant with respect to rotations of the coordinate system, Δx0 ψ(y)dy is invariant with respect to the choice of a spherical coordinate system.  Note that for m = 2 this theorem is obvious because Δx0 =

d2 dθ2

(3.2.2.1)

and it does not depend on translations with respect to θ. We define the operator Δx0 on f ∈ D (S1 ) by Δx0 f, ψ := f, Δx0 ψ, ψ ∈ D(S1 ) in a fixed spherical coordinate system. The definition is correct. Indeed, suppose in a fixed system supp ψ ⊂ S1 \{θj = 0; π : j = 1, 2, . . . , m − 2}.

(3.2.2.2)

Then all the coefficients of Δx0 are infinitely differentiable and Δx0 ψ(y) ∈ D(S1 ). By Theorem 3.2.2.1 we obtain that the condition of Theorem 2.3.5.2 (D on Sphere) are fulfilled. Note that for m = 2 the operator Δx0 is realized by the formula (3.2.2.1) on functions of the form f = f (eiθ ), i.e., on 2π-periodic functions.

3.2. Indicators

89

Theorem 3.2.2.2 (Subsphericality of Indicator) One has [Δx0 + ρ(ρ + m − 2)]h(y, u) := s > 0

(3.2.2.3)



in D (S1 ). I.e., s is a measure on S1 . Proof. It is sufficient to prove this locally, in any spherical system. Let R(r) be finite, infinitely differentiable and nonnegative in (0; ∞) and let ψ ∈ D(S1 ) be nonnegative and satisfy (3.2.2.2). Set φ(x) := R(|x|)ψ(x0 ). Using the subharmonicity of h(x, u) (h1), Theorem 3.2.1.1 and (3.2.2.2), we have  0 ≤ h(x, u)Δφ(x)dx

  1 1 ∂ m−1 ∂ ρ + Δx0 ψ(y)rm−1 dydr. r h(y, u) m−1 r = r ∂r ∂r r2 (y,r)∈S1 ×(0;∞) Transforming the last integral we obtain     ∞ 1 ∂ ∂ rρ m−1 rm−1 R(r) rm−1 dr h(y, u)ψ(y)dy h(x, u)Δφ(x)dx = r ∂r ∂r 0 S1   ∞ rρ−2 rm−1 R(r)dr h(y, u)Δx0 ψ(y)dy. (3.2.2.4) + S1

0

Integrating by parts in the first summand we obtain   ∞  ∞ 1 ∂ m−1 ∂ ρ m−1 r R(r) r r dr = R(r)ρ(ρ + m − 2)rρ+m−3 dr. rm−1 ∂r ∂r 0 0 (3.2.2.5) Substituting (3.2.2.5) into (3.2.2.4), we have  ∞  0≤ rρ+m−3 R(r)dr h(y, u)[Δx0 + ρ(ρ + m − 2)]ψ(y)dy. S1

0

Since R(r) is an arbitrarily nonnegative function,  h(y, u)[Δx0 + ρ(ρ + m − 2)]ψ(y)dy ≥ 0 S1



for arbitrary ψ.

We will call an upper semicontinuous function which satisfies (3.2.2.3) a ρsubspherical one. Now we are going to study properties of these functions. 3.2.3 We consider the case m = 2. A ρ-subspherical function for m = 2 is called ρ-trigonometrically convex (ρ-t.c.). We will obtain for such a function a representation like in Theorems 3.1.4.4, 3.1.4.5.(*,** Hadamard). Set Tρ :=

d2 + ρ2 . dφ2

90

Chapter 3. Asymptotic Behavior of Subharmonic Functions of Finite Order

Let us find a fundamental solution of this operator. Let ρ be non-integer. Let us denote as c os ρ(φ) the periodic continuation of cos ρφ from the interval (−π, π). Theorem 3.2.3.1 (Fundamental Solution of Tρ ) One has 1 Tρ cos ρ(φ − π) = δ(φ) in D (S1 ) . 2ρ sin πρ Proof. Let f ∈ D(S1 ). We have   2π  2 c os ρ(φ − π)[f + ρ f ]dφ = lim →0

0

2π−

cos ρ(φ − π)[f  + ρ2 f ]dφ. (3.2.3.1)



Integrating by parts we obtain  2π− cosρ(φ − π)[f  + ρ2 f ]dφ 

= cosρ(φ − π) f



2π− 2π− (φ)| + ρsinρ(φ − π) f (φ)| +



2π−

f (φ)Tρ cosρ(φ − π)dφ. 

However, Tρ cos ρ(φ − π) = 0 for φ ∈ ( , 2π − ). Thus the limit in (3.2.3.1) is equal to f (0)2ρ sin πρ.  Let s be a measure on the circle S1 . Set  2π c os ρ(φ − ψ − π)s(dψ). Π(φ, s) := 0

Theorem 3.2.3.2 One has Tρ Π(φ, s) = (2ρ sin πρ) s(•) in D (S1 ). The proof is the same as GPo3) in Theorem 2.5.1.1. Theorem 3.2.3.3 (Representation of ρ-t.c.f for a non-integer ρ) Let h be ρ-t.c. on S1 for non-integer ρ and let s := Tρ h. Then h(φ) =

1 Π(φ, s). 2ρ sin πρ

The proof is like in Theorem 3.1.4.4 (*Hadamard). 3.2.4 We will suppose in this item that Vt = I, m = 2, ρ is integer. Theorem 3.2.4.1 (Condition on s) Let ρ be integer, h be ρ-t.c. and Tρ h = s. Then  2π eiρφ s(dφ) = 0. (3.2.4.1) 0

3.2. Indicators

91

Proof. We have for f ∈ D(S1 ) : s, f  = Tρ h, f  = h, Tρ f . Since eiρφ ∈ D(S1 ) for integer ρ and Tρ eiρφ = 0, we have for f := eiρφ , s, eiρ•  = h, Tρ eiρ•  = 0.



Let us denote the periodic continuation of the function f (φ) := φ from the ˜ interval [0, 2π) to (−∞, ∞) as φ. Theorem 3.2.4.2 (Generalized Fundamental Solution for Tρ ) One has 

1 1 ˜ Tρ − φ sin ρφ = δ(φ) − cos ρφ 2πρ π in D (S1 ). Proof. Let φ ∈ ( , 2π − ). Then Tρ φ˜ sin ρφ = 2ρ cos ρφ because φ˜ = φ when φ ∈ ( , 2π − ). We have also (φ sin ρφ) = sin ρφ + φρ cos ρφ. Thus 



2π

2π−

φ˜ sin ρφ Tρ f dφ = lim

• sin ρ•, f  = Tρ ˜

→0

0

φ sin ρφ Tρ f dφ. 

Integrating by parts we obtain  2π− φ sin ρφ Tρ f dφ = φ sin ρφf  (φ)|2π− − f (φ)[sin ρφ + φρ cos ρφ|2π− 

 +

2π−

Tρ [φ sin ρφ]f (φ)dφ. 

Passing to the limit as → 0 and taking in account that f is periodic and continuous we obtain  2π • sin ρ•], f  = −2πρf (0) + 2ρ cos ρφf (φ)dφ Tρ [˜ 0

= −2πρf (0) + 2ρcos ρ•, f . Set

 ˆ Π(φ, s) := 0

2π

(φ − ψ) sin ρ(φ − ψ)s(dψ).



92

Chapter 3. Asymptotic Behavior of Subharmonic Functions of Finite Order

Theorem 3.2.4.3 One has ˆ s) = −2πρ s in D (S1 ) Tρ Π(•, for s that satisfies (3.2.4.1). Proof. Using Theorem 3.2.4.2 we obtain   2π ˆ s), f  = s, f  − 1 Tρ Π(•, cos ρ(• − ψ)s(dψ), f . π 0 The last integral is zero because of Theorem 3.2.4.1.



Theorem 3.2.4.4 (Representation of ρ-t.c.f. for an integer ρ) Let h be a ρ-t.c.f.for an integer ρ and Tρ h := s. Then ˆ h(φ) = ceiφ + Π(φ, s) for some complex constant c. ˆ Proof. The function H(φ) := h(φ) − Π(φ, s) satisfies the equation Tρ H = 0 in D (S1 ) because of Theorem 3.2.4.3 and it is real. Thus H(φ) = ceiφ .  3.2.5 The class T Cρ of ρ-t.c.functions has a number of properties of subharmonic functions. The function c os ρφ is continuous and φ sin ρφ is continuous for integer ρ. Therefore any ρ-t.c.f is continuous as follows from Theorem 3.2.3.3 and 3.2.4.4. Set 1 sin ρ|φ|. E(φ) := 2ρ For any interval I := (α, β)  (−π, π) this function satisfies the equality Tρ E = δ in D (α, β), where δ is the Dirac function in zero. Let GI (ψ, φ) be the Green function of Tρ for the interval I. By definition it must be symmetric with respect to φ, ψ and have the form GI (φ, ψ) :=

1 sin ρ|φ − ψ| + AI cos ρφ cos ρψ + BI sin ρφ sin ρψ, 2ρ

(3.2.5.1)

where AI , BI are chosen such that GI (φ, ψ) is equal to zero on ∂{I × I}. An explicit form of GI is given by ⎧ sin ρ(ψ−α) ⎨ sin ρ(β−φ) , for ψ < φ; ρ sin ρ(β−α) GI (φ, ψ) = ⎩ sin ρ(β−ψ) sin ρ(φ−α) , for φ < ψ. ρ sin ρ(β−α)

3.2. Indicators

93

The following assertion is analogous to the Riesz theorem (Theorem 2.6.4.3): Theorem 3.2.5.1 (Representation on I) Let h ∈ T Cρ and let I be an interval of length mes I < π/ρ. Then β h(φ) = Yρ (φ, h) −

GI (φ, ψ)s(dψ), α

where Yρ (φ, h) is the only solution of the boundary problem: Tρ Y = 0, Y (α) = h(α), Y (β) = h(β)

(3.2.5.2)

and s := Tρ h. Proof. Set

β ΠI (φ, s) :=

GI (φ, ψ)s(dψ). α

One can check as in Theorem 3.2.3.2 that Tρ ΠI = −s in D (I). Then the function Yρ (φ) := h(φ) + ΠI (φ, s) satisfies the conditions (3.2.5.2).



The explicit form of Yρ (φ) is Yρ (φ) =

h(α) sin ρ(β − φ) + h(β) sin ρ(φ − α) . sin ρ(β − φ)

(3.2.5.3)

Since ΠI (φ) ≥ 0 we have Theorem 3.2.5.2 (ρ-Trigonometric Majorant) Suppose h ∈ T Cρ and Yρ (φ) is the solution of (3.2.5.2). Then h(φ) ≤ Yρ (φ), φ ∈ I if β − α < π/ρ. This inequality can be written in the symmetric form h(α) sin ρ(β − φ) + h(φ) sin ρ(α − β) + h(β) sin ρ(φ − α) ≥ 0

(3.2.5.4)

for max(α, φ, β) − min(α, φ, β) < π/ρ. It is called the fundamental relation of indicator. Theorem 3.2.5.3 (Subharmonicity and ρ-t.c.) A function h(φ) ∈ T Cρ iff the function u(reiφ ) := h(φ)rρ is subharmonic in R2 . Proof. Sufficiency follows from Theorem 3.2.2.2. Let us prove necessity. The function u1 (z) := rρ sin ρ|φ| is subharmonic. Actually, it is harmonic for φ = 0, r = 0 and can be represented in the form u1 (z) = max(rρ sin φ, −rρ sin φ)

94

Chapter 3. Asymptotic Behavior of Subharmonic Functions of Finite Order

in a neighborhood of the line φ = 0. Hence, it is subharmonic because of sh2), Theorem 2.6.1.1 (Elementary Properties). The function u2 (z) := rρ ΠI (φ, s) is subharmonic because of sh5) and sh4), Theorem 2.6.1.1. The function rρ Yρ (φ) is harmonic for r > 0. This can be checked directly. Hence, u(z) is subharmonic for r > 0 because of Theorem 3.2.5.1. By Theorem 2.6.2.2 u(z) is also subharmonic for r = 0, because it is, obviously, continuous at z = 0.  Theorem 3.2.5.4 (Elementary Properties of ρ-t.c.Functions) One has tc1)

If h ∈ T Cρ , then Ah ∈ T Cρ for A > 0;

tc2)

If h1 , h2 ∈ T Cρ , then h1 + h2 , max(h1 , h2 ) ∈ T Cρ .

These properties follow from Theorem 3.2.5.3 and properties of subharmonic functions. Exercise 3.2.5.1 Prove Theorem 3.2.5.4. Similarly to (usual) convexity, ρ-t.convexity of functions implies several analytic properties. Theorem 3.2.5.5 Let h ∈ T Cρ ; then there exist right (h+ ) and left (h− ) derivatives and they coincide everywhere except, maybe, for a countable set of points. Proof. It is enough to prove these properties for the potential  β sin ρ|φ − ψ|s(dψ), Π(φ) := α

because of (3.2.5.1) and Theorem 3.2.5.1. We will prove that Π+ (φ)

φ−0 

β

cos ρ(φ − ψ)s(dψ) + ρμ(φ) − ρ

=ρ α

Π− (φ)

(3.2.5.5)

cos ρ(φ − ψ)s(dψ),

(3.2.5.6)

φ+0

φ−0 

β

cos ρ(φ − ψ)s(dψ) − ρμ(φ) − ρ

=ρ

cos ρ(φ − ψ)s(dψ);

α

φ+0

where μ(φ) is the measure, concentrated in the point φ. We have for Δ > 0: Π(φ + Δ) − Π(φ) = Δ

φ−0 

sin ρ|φ + Δ − ψ| − sin ρ|φ − ψ| s(dψ) Δ

α

sin ρΔ + μ(φ) + Δ

φ+Δ 

β

···+ φ+0

··· φ+Δ

.

3.2. Indicators

95

Let us estimate the second integral.We have   sin ρ|φ + Δ − ψ| − sin ρ|φ − ψ|   s(dψ) ≤ 2 sin ρΔ [s(φ + Δ) − s(φ + 0)] = o(1)    Δ Δ

φ+Δ  

φ+0

when Δ → +0. Passing to the limit, we obtain (3.2.5.5). The equality (3.2.5.6) is obtained in the same way when Δ < 0. Since μ(φ) = 0 at most in a countable set, for all the other points Π+ (φ) =  Π− (φ).  3.2.6 Now we consider the case m ≥ 3. We will obtain for the ρ-subspherical function a representation like for the ρ-trigonometrically convex functions. Theorem 3.2.6.1 (Subharmonicity and Subsphericality) Let h be subspherical in a neighborhood of y ∈ S1 . Then the function u(x) := h(y)rρ , x = ry is subharmonic in the corresponding neighborhood of the ray x = ry : 0 < r < ∞. Proof. Let f ∈ D (Rm \ 0). We can represent it in the form f := f (rx), x ∈ S1 where f (•x) ∈ D (0, ∞) for any x. Then ∞  u(rx)Δf (rx)rm−1 drdx u, Δf  = 0 S1

∞  =

u(rx)

1 ∂ m−1 ∂ r f (rx)rm−1 drdx rm−1 ∂r ∂r

0 S1

∞  +

u(rx)

1 Δx0 f (rx)rm−1 drdx. r2

0 S1

Integrating by parts in the first integral, we obtain ∞  ρ(ρ + m − 2)rρ+m−3 h(x)f (rx)drdx. 0

S1

Set Sρ := Δx0 + ρ(ρ + m − 2).

(3.2.6.1)

Together with the second summand we obtain ⎡ ⎤ ∞  u, f  = ⎣ h(x)Sρ f (rx)dx⎦ rρ+m−3 dr > 0 0

if f (rx) ≥ 0.

S1



96

Chapter 3. Asymptotic Behavior of Subharmonic Functions of Finite Order

Note that the Riesz measure for such u has the form μ(rm−1 drdx) = rρ+m−3 drνh (dx), where νh is a positive measure on S1 , that is equal to Sρ h in D (S1 ). For a non-integer ρ set ∞ Eρ (x, y) := Gp (x, ry, m)rρ+m−3 dr, 0

where Gp is the primary kernel and x, y ∈ S1 . Theorem 3.2.6.2 For non-integer ρ and any ρ-subspherical function h one has  h(x) = Eρ (x, y)νh (dy). S1

Proof. Set in (3.1.4.6) v := rρ h(x). It is clear that v ∈ U [ρ]. We have  ∞ ρ

Gp (rx, ty, m)tρ+m−3 dtνh (dx).

r h(x) = S1 0

Now we make the change t := t/r and use the homogeneity of Gp (rx, ty, m).



Exercise 3.2.6.1 Show that Eρ (x, y) is a fundamental solution of the operator Sρ . For an integer ρ = p set Eρ (x, y)

1 :=

∞ Gp−1 (x, ry)r

ρ+m−3

0

Gp (x, ry)rρ+m−3 dr.

dr + 1

Exercise 3.2.6.2 Prove the next Theorem 3.2.6.3 For any integer ρ = p and any ρ-subspherical function h one has  h(x) = Yp (x) + Eρ (x, y)νh (dy) S1

where Yp is some p-spherical function. For any p-spherical function Y ,  Y (x)νh (dy) = 0. S1

3.2.7 We return to the general case when x ∈ Rm , Vt is a one-parametric group, ρ(r) is a proximate order and u ∈ SH(ρ(r)). The following theorem represents indicators in a form of limits in the usual topology.

3.2. Indicators

97

Theorem 3.2.7.1 (Classic Indicators) One has h(x, u) = sup[lim sup utj ]∗ (x) = [lim sup ut (x)]∗ T

tj →∞

(3.2.7.1)

t→∞

where ∗ can be deleted outside a set of zero capacity, and h(x, u) = inf [lim sup utj ]∗ (x), tj →∞

T

(3.2.7.2)

where T is the set of all the sequences that tend to infinity. Proof. Let us prove (3.2.7.1). Set h(x, u, {tj }) := lim sup utj (x). tj →∞

(3.2.7.3)

Let v ∈ Fr[u] and utj → v in D . Then h∗ (x, u, {tj }) = v(x)

(3.2.7.4)

by Theorem 2.7.3.3. (H. Cartan+). Thus sup h∗ (x, u, {tj }) ≥ h(x, u).

(3.2.7.5)

T

Let > 0 be arbitrarily small, and tj := tj (x) be a sequence such that h∗ (x, u, {tj }) ≥ sup h∗ (x, u, {tj }) − . T

We can find a subsequence {tj } (we keep the same notation for it) and v ∈ Fr[u] such that utj → v in D . From (3.2.7.4) we obtain h(x, u) ≥ v(x) ≥ sup h∗ (x, u, {tj }) − . T

Thus the reverse inequality to (3.2.7.5) holds. Therefore h(x, u) = sup h∗ (x, u, {tj }). T

Let us prove the second equality in (3.2.7.1). Since sup h(x, u, {tj }) = lim sup ut (x) t→∞

T

we have

h(x, u) ≥ [lim sup ut ]∗ (x).

(3.2.7.6)

t→∞

Let us prove the opposite inequality. Let v ∈ Fr[u]. There exists a sequence tj → ∞ such that utj → v in D (Rm ). By (3.2.7.4) [lim sup ut ]∗ ≥ h∗ (x, u, {tj }) = v(x). t→∞

98

Chapter 3. Asymptotic Behavior of Subharmonic Functions of Finite Order

Since it holds for every v ∈ Fr[u] we have the reverse inequality to (3.2.7.6). Hence, (3.2.7.1) is proved completely. Let us prove (3.2.7.2). From (3.2.7.4) we have inf h∗ (x, u, {tj }) ≤ v(x) T

for all v ∈ Fr[u]. Therefore inf h∗ (x, u, {tj }) ≤ h(x, u).

(3.2.7.7)

T

Let us prove the opposite inequality. Let {tj } be any sequence that tends to ∞. Let us find a subsequence {tj  } such that utj → v in D (Rm ). Then h(x, u, {tj }) ≥ lim sup utj (x). j  →∞

Taking ∗ from the two sides of this inequality and using Theorem 2.7.3.3, we obtain h∗ (x, u, {tj }) ≥ [lim sup utj ]∗ (x) = v(x) ≥ h(x, u). j  →∞



This implies the reverse inequality to (3.2.7.7). Hence (3.2.7.2) holds. Corollary 3.2.7.2 If all the functions (3.2.7.3) are upper semicontinuous, then h(x, u) = lim sup ut (x),

h(x, u) = lim inf ut (x). t→∞

t→∞

Proof. We have h∗ (x, u, {tj }) = h(x, u, {tj }) and thus h(x, u) = sup[lim sup utj ](x) = lim inf ut (x), T

t→∞

tj →∞

h(x, u) = inf [lim sup utj ](x) = lim inf ut (x). T

t→∞

tj →∞



Theorem 3.2.7.3 (Indicators of Harmonic Function) Let u ∈ SH(ρ(r)) be harmonic for all the large |y| in a “cone” of the form CoΩ := {y = Pt x : x ∈ Ω, t ∈ (0; ∞)} where Ω ⊂ S1 . Then h(x, u) = lim sup ut (x)

(3.2.7.8)

t→∞

and h(x, u) = lim inf ut (x) t→∞

for x ∈ CoΩ .

(3.2.7.9)

3.3. Densities

99

Proof. The harmonicity of u in CoΩ implies [ut ] (x) = ut (x) for large t and sufficiently small when x ∈ CoΩ . The family [ut ] is uniformly continuous by reg3), Theorem 2.3.4.5 (Properties of Regularizations). Thus the function (3.2.7.5) is continuous. Therefore we can use Corollary 3.2.7.2.  Theorem 3.2.7.4 (Indicator for m = 2) Let u ∈ SH(R2 ). Then h(z, u) = lim sup ut (x).

(3.2.7.10)

t→∞

I.e., the star in (3.2.7.1) can be deleted. Proof. Let as denote as h1 (z, u) the right part of (3.2.7.10). The “homogeneity” of the indicator (3.2.1.10) and also of h1 (z, u) implies the following property: if the inequality h1 (z, u) < h(z.u) holds for some z0 , it holds on the whole orbit z = {Pt z0 : 0 < t < ∞} that has a positive capacity in R2 . This contradicts Theorem 3.2.7.1.



3.3 Densities 3.3.1 In the sequel G is an open set, K is a compact set and E a bounded Borel set. Let μ ∈ M(ρ(r)) and Fr[μ] := Fr[μ, ρ(r), Vt , Rm ] be the limit set of μ. Set Δ(G, μ) := sup{ν(G) : ν ∈ Fr[μ]}; Δ(E, μ) := inf{Δ(G, μ) : G ⊃ E}; Δ(K, μ) := inf{ν(K) : ν ∈ Fr[μ]}; Δ(E, μ) := sup{Δ(K) : K ⊂ E}. The quality Δ(E, μ), (Δ(E, μ)) is called the upper (lower) density of μ relative to the proximate order ρ(r) and the family Vt . Theorem 3.3.1.1 (Properties of Densities) The following properties hold: dens1) if E = ∅, then Δ(E, •) = Δ(E, •) = 0; dens2) ∀E, Δ(E, •) ≤ Δ(E, •); dens3) monotonicity: Δ, Δ(E1 , •) ≤ Δ, Δ(E2 , •) for E1 ⊂ E2 ; dens4) generalized semi-additivity1 with respect to a set:

1 See

Δ(E1 ∪ E2 , •) + Δ(E1 ∩ E2 , •) ≤ Δ(E1 , •) + Δ(E2 , •),

(3.3.1.1)

Δ(E1 ∪ E2 , •) + Δ(E1 ∩ E2 , •) ≥ Δ(E1 , •) + Δ(E2 , •);

(3.3.1.2)

Exercise 3.3.1.1

100

Chapter 3. Asymptotic Behavior of Subharmonic Functions of Finite Order

dens5) continuity from the right and from the left. En ↑ E =⇒ Δ(En , •) ↑ Δ(E, •); Kn ↓ K =⇒ Δ(Kn , •) ↓ Δ(K, •), (3.3.1.3) En ↓ E =⇒ Δ(En , •) ↓ Δ(E, •); Gn ↑ G =⇒ Δ(Gn , •) ↑ Δ(G, •); (3.3.1.4) dens6) semi-additivity and positive homogeneity with respect to μ, i.e., Δ(E, μ1 + μ2 ) ≤ Δ(E, μ1 ) + Δ(E, μ2 );

(3.3.1.5)

Δ(E, μ1 + μ2 ) ≥ Δ(E, μ1 ) + Δ(E, μ2 );

(3.3.1.6)

Δ, Δ(E, λμ) = λΔ, λΔ(E, μ)

(3.3.1.7)

for λ ≥ 0; dens7) invariance with respect to the map (•)[t] (see, 3.1.2.4a), i.e., t−ρ−m+2 Δ, Δ(Pt E, •) = Δ, Δ(E, •). Proof of Theorem 3.3.1.1. The property dens1) holds because the empty set is open by definition. The properties dens2) and dens3) hold because of the monotonicity of ν. Let us prove dens4). Since ν is a measure we have ν(G1 ∪ G2 , μ) + ν(G1 ∩ G2 , μ) = ν(G1 , μ) + ν(G2 , μ) for any G1 ⊃ E1 and G2 ⊃ E2 . From this we obtain ν(G1 ∪ G2 , μ) + ν(K1 ∩ K2 , μ) ≤ ν(G1 , μ) + ν(G2 , μ)

(3.3.1.8)

for K1 ⊂ E1 and K2 ⊂ E2 . The right side of (3.3.1.8) is no larger than Δ(G1 , •) + Δ(G1 , •). Now we can take supremum over ν ∈ Frμ in the first summand of the left side and infimum in the second summand. Thus we obtain Δ(G1 ∪ G2 , •) + Δ(K1 ∩ K2 , •) ≤ Δ(G1 , •) + Δ(G1 , •).

(3.3.1.9)

Since Δ(E, •) and Δ(E, •) are monotonic with respect to E, inf{Δ(G1 ∪ G2 , •) : G1 ⊃ E1 , G2 ⊃ E2 } = Δ(E1 ∪ E2 , •) and sup{Δ(K1 ∩ K2 , •) : K1 ⊂ E1 , K2 ⊂ E2 } = Δ(E1 ∪ E2 ). Thus we obtain the first inequality in dens4) from (3.3.1.9). The second one can be obtained analogously2. 2 Exercise

3.3.1.2

3.3. Densities

101

Let us prove dens5). For arbitrary G ⊃ K there exists n0 such that Kn ⊂ G for n > n0 . According to dens3), Δ(K, •) ≤ Δ(Kn , •) ≤ Δ(G, •). Hence, Δ(K, •) ≤ lim Δ(Kn , •) ≤ Δ(G, •). n→∞

Taking infimum over all G ⊃ K, we obtain the second assertion in (3.3.1.3). For Gn ↑ G we have the equality lim Δ(Gn , •) = sup Δ(Gn , •) = Δ(G, •)

n→∞

(3.3.1.10)

n

because one can change the order of taking the supremum on n and on ν ∈ Fr[μ]. Let En ↑ E and let be arbitrarily small. One can find Gn ⊃ En such that

Since G :=

∞ 1

Δ(Gn , •) ≤ Δ(En , •) + . Gn ⊃ E we have Δ(Gn , •) − ≤ Δ(En , •) ≤ Δ(E, •) ≤ Δ(G, •).

Using (3.3.1.10), we obtain Δ(E, •) − lim Δ(En , •) ≤ . n→∞

Since is arbitrarily small, Δ(E, •) ≤ lim Δ(En , •) n→∞

and hence the first assertion in (3.3.1.3) holds. The assertion (3.3.1.4) can be proved analogously.3 Let us prove dens6). One has Δ(G, μ1 + μ2 ) = sup{ν(G) : ν ∈ Fr[μ1 + μ2 ]}. Since Fr[μ1 + μ2 ] ⊂ Fr[μ1 ] + Fr[μ2 ] (see frmu1), Theorem 3.1.3.4 (Properties of μ → Fr[μ])) one can continue the previous equality as ≤ sup{ν(G) : ν ∈ Fr[μ1 ] + Fr[μ2 ]} = sup{ν(G) : ν ∈ Fr[μ1 ]} + sup{ν(G) : ν ∈ Fr[μ2 ]} = Δ(G, μ1 ) + Δ(G, μ2 ). Passing to the infimum over G ⊃ E, we obtain (3.3.1.5). The assertions (3.3.1.6) and (3.3.1.7) can be proved analogously.4 The properties dens7) follow from the invariance of Fr[μ] (see frm3), Theorem 3.1.3.3. (Properties of Fr[μ])).  3 See 4 See

Exercise 3.3.1.3 Exercise 3.3.1.4

102

Chapter 3. Asymptotic Behavior of Subharmonic Functions of Finite Order

Exercise 3.3.1.1 Prove the subadditivity of Δ(E, •) : Δ(E1 ∪ E2 , •) ≤ Δ(E1 , •) + Δ(E2 , •) and the superadditivity of Δ(E, •) : Δ(E1 ∪ E2 , •) ≥ Δ(E1 , •) + Δ(E2 , •) from Theorem 3.3.1.1. Exercise 3.3.1.2 Prove (3.3.1.2). Exercise 3.3.1.3 Prove (3.3.1.4). Exercise 3.3.1.4 Prove (3.3.1.6) and (3.3.1.7). Set for I ⊂ (0, ∞) and Ω ⊂ S1 , CoΩ (I) := {x = Pt y : y ∈ Ω, t ∈ I}. Also set It := (0, t). Theorem 3.3.1.2 (Cone’s Densities) One has Δ, Δ(CoΩ (It )) = tρ+m−2 Δ, Δ(CoΩ (I1 )). We obtain this from dens7), Theorem 3.3.1.1, taking E := CoΩ (I1 ). Exercise 3.3.1.5 Show that for m = 2, S1 = {|z| = 1}, Ω = {z = eiφ : φ ∈ (α, β)} CoΩ (It ) is a sector of radius t corresponding to the arc (α, β) on the unit circle. 3.3.2 Let δ(E) be a monotonic function of E ∈ Rm . A set E is called δ-squarable if sup δ(K) = inf δ(G). K⊂E

G⊃E

(3.3.2.1)

Example 3.3.2.1 Let δ(E) be a measure. Then (3.3.2.1) implies δ(∂E) = 0, i.e., E is δ-squarable in the sense of Section 2.2.3. Exercise 3.3.2.2 Prove the next Theorem 3.3.2.1 If Δ(∂E) = 0, then E is Δ-squarable. If E is Δ-squarable, then Δ(∂E) = 0. Set Et := {x : ∃y ∈ E : |x − y| < t}. This is a t-extension of E.

3.3. Densities

103

A family of sets A1 is said to be dense in a family A2 if for each set E2 ∈ A2 and an arbitrarily small > 0 there exists a set E1 ∈ A1 such that E1 ΔE2 := (E1 \ E2 ) ∪ (E2 \ E1 ) ⊂ (∂E2 ) .

(3.3.2.2)

Exercise 3.3.2.3 Prove Theorem 3.3.2.2 The relation “to be dense in” is reflexive and transitive. I.e., A1 is dense in A1 , and {A1 is dense in A2 } ∧ {A2 is dense in A3 } =⇒ {A1 is dense in A3 }.

(3.3.2.3)

There are lots of squarable sets. Theorem 3.3.2.3 For any monotonic δ(E) the class of δ-squarable sets is dense in the class of all the subsets of Rm . Proof. For any E ⊂ Rm set E(t) := E ∪ (∂E)t .

(3.3.2.4)

EΔE(t1 ) ⊂ (∂E)t2

(3.3.2.5)

One can check that and

◦

E(t1 ) ⊂ E(t2 ) for t1 < t2 .

(3.3.2.6)

◦

The function f (t) := δ(E(t)) is monotonic. Hence, its set of continuity points has a concentration point at t = 0. Suppose > 0 is arbitrarily small, and t0 < is a continuity point for f (t). From (3.3.2.6) we have ◦

lim δ(E(t)) ≤ sup δ(K) ≤ inf δ(G) ≤ lim δ(E(t)).

t→t0 −

K⊂Et0

G⊃Et0

t→t0 +

Hence, Et0 is δ-squarable. From (3.3.2.5) we have EΔE(t0 ) ⊂ (∂E) . Set



cl

Δ (E) = lim sup μt (E); Δcl (E) = lim inf μt (E). t→∞

t→∞

These are classic densities determined without D -topology. They are monotonic.

104

Chapter 3. Asymptotic Behavior of Subharmonic Functions of Finite Order

The following assertion connects these densities to Δ and Δ. cl

Theorem 3.3.2.4 (Classic Densities) For any Δ -squarable set E, cl

Δ (E) = sup{ν(E) : ν ∈ Fr[μ]} = Δ(E, μ).

(3.3.2.7)

cl

For any Δ -squarable set E, Δcl (E) = inf{ν(E) : ν ∈ Fr[μ]} = Δ(E, μ).

(3.3.2.7 )

The theorem follows obviously from the following assertion. Theorem 3.3.2.5 One has cl

cl

sup Δ (K) ≤ sup ν(E) ≤ Δ(E) ≤ inf Δ (G); K⊂E

G⊃E

ν∈Fr

sup Δcl (K) ≤ inf ν(E) ≤ Δ(E) ≤ inf Δcl (G). K⊂E

ν∈Fr

G⊃E

(3.3.2.8) (3.3.2.9)

Proof. Let us prove, for example, (3.3.2.9). Let us choose any G and K such that K ⊂ E ⊂ G. We can find a sequence tj → ∞ such that lim μtj (G) = Δcl (G).

j→∞

Choose a subsequence tjn such that μtjn → ν in D for some ν ∈ Fr. Using Theorems 2.3.4.4.(D and C*) and 2.2.3.1.(C*-limits), we obtain ν(G) ≤ lim inf μtjn (G) = Δcl (G).

(3.3.2.10)

Δcl (K) ≤ lim sup μtjn (K) ≤ ν(K).

(3.3.2.11)

n→∞

By the same theorems n→∞

From (3.3.2.10) and (3.3.2.11) we obtain Δcl (K) ≤ ν(E) ≤ ν(G) ≤ Δcl (G)

(3.3.2.12)

because of monotonicity of ν(E). Taking supremum over all K ⊂ E and infimum over all G ⊃ E, we obtain (3.3.2.9).  Exercise 3.3.2.4 Prove (3.3.2.8). Corollary 3.3.2.6 The following holds: cl

Δ (Kt ) = Δ(Kt , μ) = tρ+m−2 Δ(K1 , μ), t ≥ 0,

(3.3.2.13)

Δ(K1 , μ), t ≥ 0.

(3.3.2.13 )

cl

Δ (Kt ) = Δ(Kt , μ) = t

ρ+m−2

where Kt = {x : |x| < t} is the ball. Proof. The right equalities follow from Theorem 3.3.1.2 with Ω := S1 . The left equalities hold at least for one t because of Theorem 3.3.2.4 and hence for all t. 

3.3. Densities

105

3.3.3 Let us note generally speaking that values of Δ and Δ on the sets CoΩ (It ) do not determine their values even on the sets CoΩ (I) for I = (t1 , t2 ). However the following assertion holds. Theorem 3.3.3.1 (Existence of Density) Let Φ be a dense ring (see, 2.2.3) on S1 . Then the conditions Δ(CoΩ (It )) = Δ(CoΩ (It )) (3.3.3.1) for Ω ∈ Φ and some t determine uniquely a measure Δ(Ω) on S1 . Fr[μ] consists of one single measure ν and ν(CoΩ (It )) = tρ+m−2 Δ(Ω)

(3.3.3.2)

for all the t ∈ (0, ∞). To prove this we need an assertion that is valuable by itself. Set Δ(Ω) := Δ(CoΩ (I1 )); Δ(Ω) := Δ(CoΩ (I1 )) for Ω ∈ S1 .

(3.3.3.3)

We will call them angular densities because for m = 2 and Vt ≡ I, Ω determines an angle in the plane. Let ΩG denote an open set in S1 and ΩK a closed one. Theorem 3.3.3.2 (Angular Densities) One has Δ(Ω) = inf Δ(ΩG ); Δ(Ω) = sup Δ(ΩK ). ΩG ⊃Ω

(3.3.3.4)

ΩK ⊂Ω

Proof. We need to prove two assertions: ∀ > 0 ∃ΩG : Δ(ΩG ) < Δ(Ω) + ;

(3.3.3.5)

: Δ(Ω ) > Δ(Ω) − .

(3.3.3.6)

∀ > 0 ∃Ω

K

K

Let us prove (3.3.3.5). Set ΩG ( ) := CoΩ (I1+ ) ∪ {|x| < }. This is an open set that contains CoΩ (I1 ). One can show the following: Exercise 3.3.3.1 For every open set G ⊃ CoΩ (I1 ) there exists > 0 and ΩG ⊂ S1 such that ΩG ( ) ⊂ G. We will show Δ(ΩG ( )) < Δ(ΩG ) + o(1)

(3.3.3.7)

uniformly with respect to Ω ⊂ S1 while → 0. We have from Exercise 3.3.1.1, G

Δ(ΩG ( )) ≤ Δ(CoΩ (I1+ )) + Δ({|x| < }).

(3.3.3.8)

106

Chapter 3. Asymptotic Behavior of Subharmonic Functions of Finite Order

The property dens7), Theorem 3.3.1.1, gives Δ(CoΩG (I1+ )) = Δ(CoΩG (I1 ))(1 + )ρ+m−2 . Since Δ(CoΩG (I1 )) ≤ Δ({|x| < 1}) we have Δ(CoΩG (I1+ )) = Δ(CoΩG (I1 )) + o(1)

(3.3.3.9)

uniformly with respect to ΩG ⊂ S1 as → 0. By dens7) we also have Δ({|x| < }) = Δ({|x| < 1}) ρ+m−2 = o(1).

(3.3.3.10)

From (3.3.3.10), (3.3.3.9) and (3.3.3.8) we obtain (3.3.3.7). Hence (3.3.3.5) is proved. Let us prove (3.3.3.6). Set ΩK ( ) := CoΩK (I¯1− ) \ {|x| < } where I¯ is the closure of I. One can show the following: Exercise 3.3.3.2 For any compact K ⊂ CoΩ (I1 ) there exist ΩK ⊂ Ω and > 0 such that K ⊂ ΩK ( ) ⊂ CoΩ (I1 ). From the definition of Δ(Ω) and the monotonicity we obtain (3.3.3.6).



Proof of Theorem 3.3.3.1. Suppose (3.3.3.1) holds. The property dens7), Theorem 3.3.1.1, implies (3.3.3.1) for all the t ∈ (0, ∞). Set Δ(Ω) := Δ(Ω) = Δ(Ω) for Ω ∈ Φ. Let us prove that Δ satisfies the conditions Δ1)–Δ3) from Section 2.2.3. The conditions Δ1) and Δ2) follow from dens3) and dens4), Theorem 3.3.1.1, Exercise 3.3.1.1. Let us prove Δ3). By Theorem 3.3.3.2 for arbitrary Ω ∈ Φ and > 0 we can choose ΩG ⊃ Ω such that Δ(Ω) > Δ(ΩG ) − and ΩK ⊂ Ω such that Δ(Ω) < Δ(ΩK ) + . Suppose Ω ∈ Φ satisfies the condition ΩK ⊂ Ω ⊂ ΩG . Then Δ(Ω ) = Δ(Ω ) ≤ Δ(ΩG ) ≤ Δ(Ω) + = Δ(Ω) + and implying Δ3).

Δ(Ω) − = Δ(Ω) − ≤ Δ(ΩK ) ≤ Δ(Ω ) = Δ(Ω ), 

Chapter 4

Structure of Limit Sets 4.1 Dynamical systems 4.1.1 The most complete and effective description of an arbitrary limit set can be done in terms of dynamical systems (see, [An]). A family of the form T t : M → M, t ∈ R, on a compact metric space (M, d) with a metric d(•, •) is a dynamical system (T • , M ) if it satisfies the condition T t+τ = T t ◦ T τ , t, τ ∈ R and the map (t, m) → T t m is continuous with respect to (t, m), for all t ∈ R, m ∈ M. Let m, m ∈ M , and , s > 0. An ( , s)-chain from m to m is a finite sequence m0 = m, m1 , . . . , mn = m , satisfying the conditions d(T tj mj , mj+1 ) < , j = 0, 1, . . . , n − 1, for some tj ≥ s. A dynamical system (T • , M ) is called chain recurrent (see, [HS]), if for an arbitrarily small > 0 and an arbitrarily large s > 0 there exists an ( , s)-chain in M from m to m. Theorem 4.1.1.1 (Properties of Chain Recurrence) Let (T • , M ) be a dynamical system on a compact set. Then the following conditions are equivalent: cr1)

M is connected and (T • , M ) is chain recurrent;

cr2)

for every open proper U ⊂ M satisfying T t U ⊂ U, −∞ < t < 0,

(4.1.1.1)

the boundary ∂U contains a nonempty T • -invariant subset of M ;

108

cr3)

Chapter 4. Structure of Limit Sets

for every closed proper K ⊂ M satisfying T t K ⊂ K, t ≥ 0,

(4.1.1.2)

the boundary ∂K contains a nonempty T • -invariant subset of M ; cr4)

there does not exist any open proper V ⊂ M satisfying T τ clos V ⊂ V for some τ > 0;

cr5)

for any small > 0, large s > 0, and every pair of points m, m there exists an ( , s)-chain from m to m .

Proof. The conditions cr2) and cr3) are equivalent. Let us prove, for example, cr2)=⇒ cr3). Set U := M \ K. It is open. Applying to (4.1.1.2) T −t and, using the invariance of M , we obtain (4.1.1.1) for U . Hence ∂U contains a nonempty invariant subset of M . Since ∂K = ∂U we obtain cr2). Let us prove the implication cr1)=⇒cr3). Let K ⊂ M be closed, proper and satisfy (4.1.1.2). Since M is proper ∂K is nonempty. Let W denote the interior of K in M . The continuity of T and (4.1.1.2) imply T tW ⊂ W

(4.1.1.3)

for t ≥ 0. Indeed, T t W ⊂ K. It must be open. Thus it cannot contain any point of ∂K, since else it would contain some neighborhood of this point, contradicting the definition of ∂K. Suppose that ∂K does not contain any nonempty T -invariant set. Let us show that there exists s > 0 such that T s K ⊂ W.

(4.1.1.4)

For any m ∈ ∂K there exists t = t(m) such that T t m ∈ W . There exists a neighborhood Vm of m in ∂K that passes to W under T t(m) -action because of continuity of T t m on m. We also have T t Vm ⊂ W for t > t(m) because of (4.1.1.3). Since ∂K is compact we can cover it by a finite number of neighborhoods and obtain s such that (4.1.1.5) T s ∂K ⊂ W. (4.1.1.5) and (4.1.1.3) give (4.1.1.4). Set := 0.5d(∂K, T s K). From (4.1.1.2) we see that T t K ⊂ T s K for t > s. Therefore there does not exist any ( , s)-chain from a small neighborhood of a point m ∈ ∂K to itself. This contradicts the chain recurrence of M . Let us prove cr3)=⇒cr4). Assume that there exists an open proper V ⊂ M satisfying T τ clos V ⊂ V for some τ > 0.  T t V and We will construct K that does not satisfy cr3). Set W := 0≤t≤τ

K := clos W .

4.1. Dynamical systems

109

Then T s W ⊂ W, ∀s ≥ 0. 

(4.1.1.6)



Indeed, let s = kτ + s , s ∈ [0, τ ), k ∈ Z. Then  T sW = T t+s V.

(4.1.1.7)

t∈[0,τ ]

Since T τ V ⊂ V we have T t+kτ V ⊂ T t V for t > 0. From (4.1.1.7) we obtain       T t+s +kτ V ⊂ T t+s V = Tt V T sW = t∈[0,τ ]

=



T tV ∪

t∈[s ,τ ]



t ∈[s ,τ +s ]

t∈[0,τ ]

T t V := W1 ∪ W2 .

t∈[τ,τ +s ]

Further we have W1 ⊂ W by definition. W2 can be represented in the form  W2 = T t+τ V. t∈[0,s ]

Since T t+τ V = T t T τ V

and

TτV ⊂ V

by the assumption we get: W2 ⊂ W1 ⊂ W. This implies (4.1.1.6). The same holds for K because of continuity of T t , i.e., K satisfies (4.1.1.2). Let us prove the equality  T t clos V. (4.1.1.8) K= 0≤t≤τ

Denote as K  the right side of (4.1.1.8). The set K  is closed because of compactness of [0, τ ]. Indeed, let the sequence tj {T vj : j = 1, 2, . . . } ∈ T tj (clos V ) converge to w. Choose a subsequence tjk → s ∈ [0, τ ]. Then v := lim vjk = lim T −tjk w = T −s w. k→∞

k→∞

Since clos V is closed, v ∈ clos V . Thus w = T s v for some s ∈ [0, τ ] and some v ∈ clos V , i.e., w ∈ K  . Now, W ⊂ K  because T t V ⊂ clos T t V = T t clos V. Hence,

K := clos W ⊂ clos K  = K  .

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Chapter 4. Structure of Limit Sets

We also have (T t V ⊂ W ∀t ∈ [0, τ ]) =⇒ (clos T t V = T t clos V ⊂ clos W = K, ∀t ∈ [0, τ ]). Hence, K  ⊂ K. Therefore K = K  , i.e., (4.1.1.8) holds. From (4.1.1.8) and T τ clos V ⊂ V we obtain T τ clos W ⊂ W . Hence T τ ∂K ⊂ W . This and ∂K ∩ W = ∅ imply T τ ∂K ∩ ∂K = ∅.

(4.1.1.9)

To obtain a contradiction and complete the proof of cr3) =⇒ cr4) we have to show that K is a proper subset, because both cases: ∂K = ∅ and ∂K = ∅ will contradict cr3). Since V is proper T t V is proper for any t ∈ (−∞, ∞). Otherwise T t V = M implies V = T −t M = M , which is a contradiction. Since V is a neighborhood of the compact set T τ clos V we can find α > 0 such that T t ◦ T τ clos V ⊂ V for t ∈ [0, α]. Then T t clos V ⊂ T −τ V for t ∈ [0, α]. By iteration of this inclusion we obtain T jt clos V ⊂ T −jτ V for any integer j. When jα > τ it follows that K ⊂ T −jτ V . The last set is proper because we mentioned already that T t V is proper for any t ∈ (−∞, ∞). Hence K is proper. So K satisfies the conditions of cr3) but ∂K does not contain a nonempty T • -invariant set. This contradiction proves the implication cr3) =⇒ cr4). Let us prove cr4) =⇒ cr5). Let > 0 be small and s > 0 be large. Let V denote the set of all m ∈ M such that there exists an ( , s)-chain from m to m . This set is open and closed. Indeed, let m ∈ V . There exists an ( , s)-chain m = m0 , . . . , mn−1 , mn = m from m to m . Choose 1 < − d(mn , mn−1 ) and consider the closed neighborhood W := {m : d(m , m ) ≤ 1 }. Then for any m ∈ W the chain m = m0 , . . . , mn−1 , mn = m is an ( , s)-chain from m to m . Hence, with every point, V contains its closed neighborhood. Therefore it is open and closed. Therefore it is a connected component of M . We also have T s m ∈ V because for that case n = 1, m0 = m, m1 = T s m. Hence T s clos V ⊂ V . If V does not coincide with the whole M the latter contradicts cr4). Hence V = M . Finally, let us prove cr5) =⇒ cr1). If M is a union of two nonempty disjoint sets A and B, then both of them are open and closed. Since M is compact, the distance between A and B is positive . Hence every ( /2, s)-chain starting at a point of A remains in A, contradicting cr5). Since for every point m ∈ M the set V from the proof of cr4)=⇒ cr5) coincides with M , cr1) holds.  • • Theorem 4.1.1.2 Let  T be chain recurrent on Mα , α ∈ A. Then T is chain recurrent on M = α∈A Mα .

This is because every ( , s)-chain from m to m in Mα is also ( , s)-chain in M .

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111

4.1.2 Here we prove two auxiliary assertions that will be used further. Theorem 4.1.2.1 Let T • be chain recurrent on a connected compact M and let {qj } be a sequence in M . Then there exist sequences {αν } and {ων } of real numbers and a sequence {pν } in M having {qj } as a subsequence, such that αν → −∞; ων → ∞

(4.1.2.1)

d(T ων pν , T αν+1 pν+1 ) → 0

(4.1.2.2)

and as ν → ∞. Proof. In addition to {αν }, {ων } and {pν } we define, by induction, a sequence { ν } of positive real numbers, tending to zero, and an increasing sequence {νj } of positive integers, such that {pνj } = {qj } and d(T ων pν , T αν+1 pν+1 ) < ν , ν = 1, 2, . . . .

(4.1.2.3)

We start by setting α1 = −1, 1 = 1, ν1 = 1, ω1 = 5 and p1 = q1 . Assume now that αν , ν , ων and pν have been chosen for ν = 1, 2, . . . , νj . Set α = ανj − 1, = νj /2, ω = ωνj .

(4.1.2.4)

By Theorem 4.1.1.1, cr5) there exists a sequence r0 := T ω qj , r1 , . . . , rm := T α qj+1 such that d(T tk rk , rk+1 ) < for k = 0, 1, . . . , m − 1, where tk ≥ ω. Now we set νj+1 = νj + m + 1. For ν = νj + k + 1, k = 0, 1, . . . , m − 1, we set αν = −tk /2, ων = tk /2, pν = T tk /2 rk , and finally, for ν = νj+1 we set αν = α, ν = , ων = ω +1, pν = qj+1 . Let us check that with this setting the properties (4.1.2.1) hold . Since ωνj+1 = ωνj + 1 we have ωνj → ∞ as j → ∞. From tk ≥ ω = ωνj we obtain αν → −∞ and ων → ∞. Hence (4.1.2.1) holds. One can see from (4.1.2.4) that ν = νj /2 → 0. To prove (4.1.2.2) it is enough to check (4.1.2.3). For k = 0 we have pν = T t0 /2 r0 = T (t0 /2)+ω qj = T (t0 /2)+ω pνj . Hence, T αν pν = T ω pνj = T ωνj pνj . Thus d(T ωνj pνj , T αν pν ) = 0 for this case. For k = 1, . . . , m − 2 and the corresponding ν we have T ων pν = T tk /2 ◦ T tk /2 rk = T tk rk

(4.1.2.5)

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Chapter 4. Structure of Limit Sets

and T αν+1 pν+1 = T −tk+1 /2 ◦ T tk+1 /2 rk+1 = rk+1 . Hence, d(T ων pν , T αν+1 pν+1 ) = d(T tk rk , rk+1 ) < = ν .

(4.1.2.6)

Finally, for the last link of the chain we obtain k = m − 1, ν = νj + m, ν + 1 = νj+1 , ανj+1 = α, T αν+1 pν+1 = T ανj+1 pνj+1 = T α qj+1 = rm . Thus (4.1.2.6) holds for k = m − 1. Hence, (4.1.2.3) also holds. Therefore (4.1.2.2) holds.  Lemma 4.1.2.2 Let pk , qk ∈ M and d(pk , qk ) → 0 as k → ∞. Then there exists a sequence {γk ↑ ∞} such that d(T τ pk , T τ qk ) → 0

(4.1.2.7)

uniformly with respect to τ ∈ [−γk+1 , γk ]. Proof. Let [−γ, γ] be a fixed segment. Then d(T τ pk , T τ qk ) → 0 uniformly in this segment. Indeed, suppose there exist sequences τj , kj such that d(T τj pkj , T τj qkj ) ≥ > 0. Choosing a subsequence we can assume that τj → τ ∈ [−γ, γ], pkj → p ∈ M and qkj → q = p. Using continuity of T τ m on (τ, m) and continuity of d(•, •) in both arguments we obtain 0 = d(p, p) ≥ > 0. This is impossible. Denote

(γ, k) := max d(T τ pk , T τ qk ). τ ∈[−γ,γ]

This function increases monotonically in γ and tends to zero for any γ as k → ∞. Choose ln such that (n, k) ≤ 1/n for k ≥ ln . Set γk+1 := n for ln < k ≤ ln+1 . One can see that (γk+1 , k) → 0 as k → ∞. Since max

τ ∈[−γk+1,γk ]

{γk } satisfies (4.1.2.7).

d(T τ pk , T τ qk ) ≤ (γk+1 , k), 

4.1.3 We connect the property of being chain recurrent with other well-known characteristics of dynamical systems ([AGL]). A point m0 ∈ M is called non-wandering (see [An]) if for any neighborhood O of m0 and arbitrarily large number s ∈ R there exists m ∈ O and t ≥ s such that T t m ∈ O.

4.1. Dynamical systems

113

This means that the “returns” take place to an arbitrarily small neighborhood of the point m0 . We shall denote as Ω(T • ) the set of non-wandering points. It is a closed invariant subset of M . The set A ⊂ M is called an attractor if it satisfies the following conditions: attr1) for any neighborhood O ⊃ A there exists a neighborhood O , A ⊂ O ⊂ O such that T t O ⊂ O t ∈ R, where T t O is the image of O ; attr2) there exists a neighborhood O ⊃ A such that T t m → A when t → ∞ for m ∈ O. Theorem 4.1.3.1 If Ω(T • ) = M, then (T • , M ) is chain recurrent; if (T • , M ) has an attractor A = M , it is not chain recurrent. Proof. The property Ω(T • ) = M obviously implies the chain recurrence for m = 1.Suppose there exists an attractor A = M . Take a point m0 that does not belong to A and choose a neighborhood O ⊃ A such that d(m0 , clos O) = 2 > 0. This is possible because an attractor is closed. Let O be chosen by attr1) and s be such that T s m ∈ O . Then there does not exist any ( , s)-chain from a small neighborhood of m0 itself. By definition (T t , M ) is not chain recurrent.  Let us give examples of dynamical systems on connected compacts that are chain recurrent. Theorem 4.1.3.2 Let M be a connected compact and let T t (−∞ < t < ∞) be the identity map. Then (T • , M ) is chain recurrent. This theorem, of course, is trivial. However, if M consists of a single point this dynamical system determines an important class of subharmonic and entire functions of completely regular growth (see [Le, Ch. III]). Let m ∈ M . Set C(m) := clos{T t m : −∞ < t < ∞}.

(4.1.3.1)

It is closed, connected and invariant. Exercise 4.1.3.1 Prove this. Let us denote as Ω(m) the set of all limits of the form Ω(m) := {m ∈ M : (∃tk → ∞)(m = lim T tk m}. k→∞

(4.1.3.2)

This is a limit set as t → ∞. It is the “tangle” at the end of the curve. Denote by A(m) the analogous set for t → −∞. Exercise 4.1.3.2 Prove that A(m) and Ω(m) are invariant. Theorem 4.1.3.3 (T • , C(m)) is chain recurrent iff A(m) ∩ Ω(m) = ∅.

(4.1.3.3)

114

Chapter 4. Structure of Limit Sets

Proof. Suppose B := A(m)∩Ω(m) = ∅. Then Ω(m) is an attractor and (T • , C(m)) is not chain recurrent by Theorem 4.1.3.1. Suppose B = ∅. We will use cr2) from Theorem 4.1.1.1. Let U be an open proper subset of C(m) satisfying (4.1.1.1). Consider two cases: i) B contains a point of U . Thus U contains a sequence of form T tk m, tk → ∞. From (4.1.1.1) we obtain that U contains T t m for all t ∈ (−∞, ∞). Thus U ⊃ C(m) and clos U = C(m). Set K = C(m) \ U . One can show that K satisfies (4.1.1.2)(see the beginning of proof of Theorem 4.1.1.1). Hence K contains the set * T tK (4.1.3.4) K ∗ := t≥0

that is invariant (Exercise 4.1.3.3). Therefore K ∗ ⊂ K ⊂ clos U \ U = ∂U . By cr2) (T t , C(m)) is chain recurrent. ii) B contains no point of U . Then B ⊂ A(m) ⊂ ∂U . By cr2) (T • , C(m)) is chain recurrent.  Exercise 4.1.3.3 Let U satisfy (4.1.1.1) and K := M \ U . Prove that K ∗ from (4.1.3.4) is invariant. 4.1.4 The connectedness of M is a necessary condition for a dynamical system to be chain recurrent. Let M be a subset of a linear space. The set M is called polygonally connected if every pair of points m1 , m2 can be connected by a polygonal path. Of course, polygonal connectedness implies connectedness and even arcwise connectedness. Theorem 4.1.4.1 Let (T • , M ) be a dynamical system such that M is a polygonally connected set. Then (T • , M ) is chain recurrent. Proof. Let U be an open proper subset of M , satisfying (4.1.1.1). We choose m1 ∈ U and m2 in an invariant subset K ∗ of K := M \ U . Then there exists a polygonal path from m1 to m2 : mθ := (j + 1 − θ)mj + (θ − j)mj+1 , for θ ∈ [j, j + 1], j = 0, 1, . . . , l − 1; m0 := m1 , ml := m2 . Now M is invariant, so for each t the continuous path θ → T t mθ lies in M . If t ∈ (−∞, 0) its initial point T t m1 belongs to U and its endpoint T t m2 belongs to K ∗ ⊂ K. For each t ∈ (−∞, 0) we set θ(t) := min[θ ∈ [0; l] : T t mθ ∈ K].

4.1. Dynamical systems

115

Then θ(t) > 0, T t mθ(t) ∈ ∂U and (4.1.1.1) implies that t → θ(t) is a decreasing function. Hence the limit θ(−∞) := lim θ(t) t→−∞

exists and is positive. Set m3 := mθ(−∞) . We claim that A(m3 ) ⊂ ∂U (A(·) is a set defined before Theorem 4.1.3.3). If θ(−∞) ∈ (j, j + 1] for some j ∈ [0, l] then θ(t) ∈ (j, j + 1] for t that is near to −∞, and T t m3 = T t mθ(t) + (θ(t) − θ(−∞))T t mj + (θ(−∞) − θ(t))T t mj+1 . The first term in the right-hand side lies in ∂U . The set M is compact and invariant so the other terms tend to zero as t → −∞. Hence A(m3 ) ⊂ ∂U . Thus ∂U contains this invariant subset and (T • , M ) is chain recurrent by cr2), Theorem 4.1.1.1.  We have the obvious Corollary 4.1.4.2 Let (T • , M ) be a dynamical system such that M is a compact convex set. Then (T • , M ) is chain recurrent. This is because the polygonal path can be taken as a line segment connecting every pair of points. 4.1.5 Let U [ρ, σ] be a set of subharmonic functions defined in (3.1.2.4). It is invariant with respect to the transformation (•)[t] defined in (3.1.2.4a). Set (subindex!) Tt v := v[et ] . (4.1.5.1) Since (•)[t] has the property (3.1.2.4b) Tt+τ v = (Tt ◦ Tτ )v, ∀t, τ ∈ R.

(4.1.5.2)

By Theorem 3.1.2.3, T• is continuous in the appropriate topology and hence (T• , U [ρ, σ]) is a dynamical system. Theorem 4.1.5.1 (Universality of U [ρ, σ]) Let (T • , M ) be a chain recurrent dynamical system on a compact set M . Then for any ρ, σ there exists U ⊂ U [ρ, σ] and a homeomorphism imb : M → U such that imb ◦T t = Tt ◦ imb, t ∈ (−∞, ∞). I.e., any dynamical system can be imbedded in (T• , U [ρ, σ]). This plot is developed in [Az(2008)]. It is sufficient to prove the theorem by supposition Pt x = tx because (T•P , U [ρ, σ]) is a dynamical system for any Pt and imb : (T• , U [ρ, σ]) → (T•P , U [ρ, σ]) where imb : u(x) → T−t TtP u(x) is also a homeomorphism of dynamical systems. Exercise 4.1.5.1 Consider Theorem 3.1.6.1 from this point of view.

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Chapter 4. Structure of Limit Sets

We need some auxiliary definitions and results. Let us denote as M(S m−1 ) the set of measures ν with bounded full variation on the unit sphere S m−1 . Introduce the metric d(ν, 0) := Var ν and consider the set K := {ν : ν > 0, d(ν, 0) ≤ 1}, i.e., the intersection of the cone of positive measures with the unit ball. The following assertion is a corollary of Keller’s theorem (see, e.g., [BP, Thm. 3.1, p. 100]). Theorem 4.1.5.2 (Imbedding) Every metric compact set can be homeomorphically imbedded to K. Thus we can assume below that for any m ∈ M there exists a positive measure Y (•, m) = Y (dx0 , m) ∈ K such that (Y (•, m1 ) = Y (•, m2 )) =⇒ (m1 = m2 )

(4.1.5.3)

and Y (•, m) is continuous with respect to the metrics. We also introduce a new coordinate system. For x := ey x0 ∈ Rm \ 0 set Pol(x) = (y, x0 ). This formula gives a one-to-one map from Rm \0 onto the cylinder Cyl := (−∞, ∞) × S m−1 . Thus, for any (y, x0 ) ∈ Cyl, Pol−1 (y, x0 ) = ey x0 . For m = 2 this is a common cylinder. 4.1.6 Proof of Theorem 4.1.5.1. We consider separately the cases of integer and noninteger ρ. Let ρ be non-integer and σ > 0. For any v ∈ U [ρ, σ], one has the representation of Theorem 3.1.4.4 (*Hadamard), v(x) = Π(x, μ, ρ)

(4.1.6.1)

where μ ∈ M[ρ, Δ] and Δ depends only on σ (Theorem 2.8.3.3). Vice versa, every μ ∈ M[ρ, Δ] generates v by (4.1.6.1) and v[t] (x) = Π(x, μ[t] , ρ). Let us “transplant” μ in Cyl. For μ that has a dense fμ (rx0 ), we set ν(dy ⊗ dx0 ) := fμ (ey x0 )e(−ρ−2)y (dy ⊗ dx0 ), i.e., the density fν of ν is defined by fν (x0 , y) := fμ (ey x0 )e(−ρ−2)y . Respectively, fμ (x0 , r) = fν (x0 , log r)rρ+2 . We can extend this equality for all μ ∈ M[ρ, Δ] by using a limit process in D topology.

4.1. Dynamical systems

117

Exercise 4.1.6.3 Do that using, for example, Theorem 2.3.4.5 (Properties of Regularization). We can also define ν as a distribution in D (Cyl). Namely, for ψ ∈ D(Cyl) we set ψ ∗ (x0 , r) := ψ(Pol−1 (x0 , log r))r−ρ−m+2 

and ν, ψ :=

ψ ∗ (x0 , r)μ(dx0 ⊗ rm−1 dr).

Exercise 4.1.6.4 Check that this definition gives the same ν. The transformation Pt x = (x0 , tr), rx0 ∈ Rm \ 0 passes to Pol ◦Pt ◦ Pol−1 (x0 , y) = (x0 , y + log t). Thus Teτ μ gives a transformation Sτ ν defined by St fν (x0 , y) := fν (x0 , y + t) for densities or by

 St ν, ψ :=

ψ(x0 , y − t)ν(dx0 ⊗ dy)

(4.1.6.2)

for distributions (ψ ∈ D(Cyl).) Exercise 4.1.6.5 Check the equivalence. From μ ∈ M[ρ, Δ] we obtain  e(ρ+m−2)y St ν(dy ⊗ dx0 ) ≤ Δ, t ∈ R.

(4.1.6.3)

y≤0

Exercise 4.1.6.6 Check this. Let X(t) be a positive function satisfying the condition ∞ X(t)dt = 1 −∞

and such that the linear hull of its translations are dense in L1 (−∞, ∞). We can choose, for example, the function 1 t2 X(t) := √ e− 2 2π because its Fourier transformation does not vanish in R (it is e−

s2 2

).

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Chapter 4. Structure of Limit Sets

Exercise 4.1.6.7 Check these properties. Let us define ν(•, m) by ⎛



ψ(x0 , y) ⎝ρ

ν(•, m), ψ :=

⎞

∞

Y (dx0 , T y−t m)X(t)dt⎠ dy.

(4.1.6.4)

−∞

(x0 ,y)∈Cyl

Now we check the property Sτ ν(•, m) = ν(•, T τ m). Using (4.1.6.2), we obtain  Sτ ν(•, m), ψ =  =

⎛ ψ(x0 , y) ⎝ρ ⎛

∞

⎞ Y (dx0 , T y+τ −tm)X(t)dt⎠ dy

−∞

ψ(x0 , y) ⎝ρ

∞

⎞ Y (dx0 , T y−t (T τ m))X(t)dt⎠ dy

−∞

= ν(•, m), T τ m. We also check the condition (4.1.6.3). 



∞ e St ν(dy ⊗ dx ) = ρy

y≤0

0

X(t)dt −∞

 eρy )ρdy

Y (dx0 , T y+τ −tm

S m−1

y≤0

∞ ≤ sup Y (S m−1 , T τ m) τ ∈R

X(t)dt ≤ 1,

−∞

since Y (•, •) ∈ K. Now we should “transplant” ν back to Rm \ 0 such that Sτ passes to (•)[eτ ] . Define μ(•, m) by (4.1.6.5) μ(•, m), ψ ∗  := ν(•, m), ψ, where ψ ∗ (rx0 ) ∈ D(Rm \ 0) and ψ(x0 , y) := ψ ∗ (ey x0 )e−(ρ−m+2)y ∈ D(Cyl). Then (μ)[eτ ] , ψ ∗  = (μ), T−τ ψ ∗  = ν, S−τ ψ = Sτ ν, ψ. The condition μ(•, m) ∈ M[ρ, σ] is also satisfied.

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119

Exercise 4.1.6.8 Check these properties. Now we use (4.1.6.1) to transplant the dynamical system to U [ρ, σ]. This completes a construction of a homomorphism (T t , M ) → (Tt , U [ρ, σ]). Let us check that it is an imbedding, i.e., we must check the one-to-one correspondence. One-to-one correspondence of v(•, m) and μ(•, m) is known (Theorem 3.1.4.4). One-to-one correspondence of μ(•, m) and ν(•, m) can be also checked easily. Exercise 4.1.6.9 Check this in detail. So we should check the one-to-one correspondence of ν(•, m) and Y (•, m). Suppose ν(•, m1 ) = ν(•, m2 ). Then ν(•, m1 ), ψ = ν(•, m2 ), ψ ∀ψ ∈ D(Cyl). In particular, set ψ(x0 , y) = φ(x0 )R(y), φ ∈ D(S m−1 ), R ∈ D(−∞, ∞). Then  ν(•, m1 ), ψ =

∞ R(y)dy

Y (•, T y−tm1 ), φS m−1 X(t)dt

−∞



= ν(•, m2 ), ψ =

∞ R(y)dy

Y (•, T y−t m2 ), φS m−1

−∞



where

(4.1.6.6)

Y (•), φS m−1 :=

φ(x0 )Y (dx0 ). S m−1

Set Fj (y) := Y (•, T y mj ), φS m−1 , j = 1, 2. From (4.1.6.6) we obtain for the convolutions (F1 ∗ X)(y) ≡ (F2 ∗ X)(y), y ∈ (−∞, ∞). Thus F1 (y) ≡ F2 (y), y ∈ (−∞, ∞) because of the property of X. Hence Y (•, T y m1 ) ≡ Y (•, T y m2 ), y ∈ (−∞, ∞).

120

Chapter 4. Structure of Limit Sets

In particular, for y = 0 we have Y (•, m1 ) = Y (•, m2 ). Hence m1 = m2 because of (4.1.5.3), and this completes the proof of one-to-one correspondence. Consider the case of an integer ρ. For this case we can use v ∈ U [ρ, σ] of the form v(x) = Π< (x, μ, ρ) + Π> (x, μ, ρ) instead of (4.1.6.1).  Exercise 4.1.6.10 Check this. 4.1.7 The most simple set satisfying the conditions of Theorem 4.1.3.3 is the set that is generated by a function v ∈ U [ρ] that has the property v[teP ] = v[t] , t ∈ (0, ∞) for some P . Then Tt+P v = Tt v, t ∈ (−∞, ∞), i.e., the dynamical system T• is periodic with the period P on the set C(v) = {Tt v : 0 ≤ t ≤ P }. Theorem 4.1.7.1 (Periodic Limit Set) For all P > 0, ρ > 0, σ > 0, there exists v ∈ U [ρ, σ] such that the dynamical system (T• , C(v)) is periodic with the period P . Proof. Suppose ρ is non-integer. Let us take μ ∈ M[ρ, Δ] such that the canonical potential Π(x, μ, [ρ]) belongs to U [ρ, σ]. This is possible because of Theorem 3.1.4.2 (*Brelot-Borel). Denote as μ∗P the restriction of μ on the spherical ring {x : 1 < |x| < eP } and set ∞  μP := TkP μ∗P . k=−∞

We have μP ∈ M[ρ, Δ] and Tt+P μP = Tt (

∞ 

T(k+1)P μ∗P ) = Tt μP .

k=−∞

Then v := Π(x, μP , [ρ]) ∈ U [ρ, σ] and Tt+P v = Tt v because of (3.1.5.0). For an integer ρ we use the function v(x) := Π< (x, μP , ρ) + Π> (x, μP , ρ).



4.2. Subharmonic function with prescribed limit set

121

4.2 Subharmonic function with prescribed limit set 4.2.1 The following two theorems describe structure of limit sets in terms of dynamical systems. Theorem 4.2.1.1 (Necessity) Let u ∈ SH(Rm , ρ, ρ(r)). Then the dynamical system (T• , Fr[u, •]) is chain recurrent. The chain recurrence is also sufficient. Theorem 4.2.1.2 (Sufficiency) Let U be a compact connected and T• -invariant subset of U [ρ, σ] for some σ > 0, such that the dynamical system (T• , U ) is chain recurrent. Then for any proximate order ρ(r) → ρ there exists u ∈ SH(Rm , ρ, ρ(r)) such that Fr[u, ρ(r), Vt , Rm ] = U. Proof of Theorem 4.2.1.1. We need the curve ut , t ≥ 1, and Fr[u, •] to be contained in a common metric space X. Thus we set X := {v ∈ SH(Rm ) : sup M (r, v)r−ρ−1 ≤ sup M (r, u)r−ρ−1 }. r≥1

r≥1

We want to use Theorem 4.1.1.1 cr 2). Let U be an open proper subset of Fr[u, •] satisfying (4.1.1.1) and let F be a T• -invariant subset of K := Fr[u, •] \ U . Such F exists. Indeed, K is closed and Tt K ⊂ K for t > 0 (see proof of Theorem 4.1.1.1, cr2)⇐⇒cr3)). Thus Ω(K) ⊂ K where Ω(•) was defined in (4.1.3.2). The set Ω(K) is invariant with respect to Tt (see Exercise 4.1.3.1). So the set of such sets F is not empty. If F intersects ∂U at a point v, then A(v) ⊂ F ∩ ∂U . Since A(v) is invariant (Exercise 4.1.3.2) ∂U contains a nonempty T• -invariant set. So we obtain the assertion of the theorem using Theorem 4.1.1.1, cr2). Suppose F does not intersect ∂U . Let U0 be an open set in X such that U0 ∩ Fr[u, •] = U, clos U0 ∩ Fr[u, •] = clos U

(4.2.1.1)

(see Exercise 4.2.1.1). Since clos U0 ∩ F = ∅ we can take a sequence of open neighborhoods U1 , U2 , . . . of F in X such that all sets clos Uj , j = 1, 2, . . . do not intersect clos U0 and Uj ↓ F . By definition of Fr[u, •] we can find intervals aj ≤ t ≤ bj with aj → ∞ such that ueaj ∈ ∂Uj , uebj ∈ ∂U0 , and uet ∈ clos U0 ∪ clos Uj for aj < t < bj . We can pass to a subsequence and assume that ueaj → w ∈ F.

(4.2.1.2)

Let us use the identity uet+aj = (ueaj )et

ρ(et )ρ(eaj ) . ρ(et+aj )

(4.2.1.3)

122

Chapter 4. Structure of Limit Sets

By (4.2.1.2), (4.2.1.3) and the property (3.1.2.2) of a proximate order we obtain uet+aj → Tt w ∈ F uniformly for any bounded set of t. Thus bj − aj → ∞. Passing to a subsequence we may assume that uebj → v ∈ Fr[u, •] ∩ ∂U0 = ∂U . Since uet+bj → Tt v and uet+bj ∈ / U0 when aj − bj < t < 0 we obtain that Tt v ∈ / U when t < 0. Hence the whole backward orbit {Tt v : t < 0} lies in ∂U , which must therefore contain the T• -invariant set A(v).  Exercise 4.2.1.1 Prove that the set  U0 := {w ∈ X : dist(v, w) < dist(v, K)/2} v∈U

satisfies the conditions (4.2.1.1). Proof. We have U0 ⊃ U =⇒ U0 ∩ Fr[u, •] ⊃ U ∩ Fr[u, •] = U. Thus U0 ∩ Fr[u, •] ⊃ U.

(4.2.1.4)

From (4.2.1.4) we have clos U0 ∩ Fr[u, •] = clos U0 ∩ clos Fr[u, •] = clos(U0 ∩ Fr[u, •]) ⊃ clos U. (4.2.1.5) Finally (4.2.1.4) ∧ (4.2.1.5) =⇒ (4.2.1.1).



4.2.2 To prove Theorem 4.2.1.2 we need some preparation. Theorems of the next Sections form the basis of the construction that we will use in the proof. Let β be an infinitely differentiable function on R such that 0 ≤ β(x) ≤ 1, β(x) = 0 for x ≤ 0 and β(x) = 1 for x ≥ 1. We can set, for example, x β(x) := A

α(y + 1)dy

−∞

where α is taken from (2.3.1.1) and ∞ α(y + 1)dy.

A= −∞

4.2. Subharmonic function with prescribed limit set

123

Suppose that the sequences {rk , σk , k = 0, 1, . . . } satisfy the following conditions: r0 = 1; rk < rk σk < rk+1 /σk+1 < rk+1 , k = 1, 2, . . . , rk+1 ↑ ∞. σk ↑ ∞; σk+1 rk σk

(4.2.2.1) (4.2.2.2)

Set 

 log r − log(rk /σk ) log r − log(rk+1 /σk+1 ) −β , log(σk rk ) − log(rk /σk ) log(σk+1 rk+1 ) − log(rk+1 /σk+1 )  log r − log(r1 /σ1 ) ψ0 (r) := 1 − β . log(σ1 r1 ) − log(r1 /σ1 )

ψk (r) := β

The sequence {ψk }, k = 0, 1, . . . forms a partition of unity with the following properties: Theorem 4.2.2.1 (Partition of Unity) One has ∞ 

ψk = 1;

(prtu1)

k=0

supp ψk ⊂ (rk /σk , rk+1 σk+1 ); ψk (r) = 1, for r ∈ (rk σk , rk+1 /σk+1 );

(prtu2) (prtu3)

supp ψk ∩ supp ψl = ∅ for |k − l| > 1;

(prtu4)

lim max ψk (r)r k→∞ r Moreover

=

lim max ψk (r)r2 k→∞ r

= 0.

max |ψk (r)r|, max |ψk (r)r2 | ≤ γk r

(prtu5)

(prtu6)

r

where γk can be made to tend to zero arbitrarily fast by choosing the sequences {σk } and {rk }. Proof. Set

 βk (r) := β

log r − log(rk /σk ) log(σk rk ) − log(rk /σk )

.

The functions βk (r) and βk+1 (r) vanish for r < rk /σk because β(x) = 0 for x ≤ 0, and both of them are equal to 1 for r ≥ σk rk because β(x) = 1 for x ≥ 1. Hence, (prtu2) holds. One has for any r ∈ (0, ∞), n 

ψk = 1 − βn+1 (r).

k=0

As mentioned, βn+1 (r) = 0 for n such that rn+1 /σn+1 > r. Thus (prtu1) holds.

124

Chapter 4. Structure of Limit Sets

Counting derivatives of ψk , we have: max |rψk (r)| ≤ r

/

0 (log(σk rk ) − log(rk /σk ))−1 + (log(σk+1 rk+1 ) − log(rk+1 /σk+1 ))−1 max |β  |(x). x

Thus we can take the right side of the inequality as γk and regulate its vanishing by choice of the ratio in (4.2.2.2). The same holds for r2 ψ  (r). Hence (prtu5) and (prtu6) are proved. 

Exercise 4.2.2.1 Check (prtu4).

4.2.3 Now we construct a function which is of zero type but has a “maximal possible” mass density. Theorem 4.2.3.1 (Maximal Mass Density Function) Let ρ(r) → ρ, ρ > 0 be a smooth proximate order (i.e., having properties (2.8.1.8)), and let γ(r), r ∈ [0, ∞), satisfy the conditions: γ(r) > 0 and γ(r) → 0, as r → ∞. Then there exists an infinitely differentiable subharmonic function Φ(x) such that (4.2.3.1) ΔΦ(x) ≥ γ(x)|x|ρ(r)−2 and (Φ)t → 0

(4.2.3.2)



in D as t → ∞. To prove Theorem 4.2.3.1 we need an elementary lemma. Theorem 4.2.3.2 (Convex Majorization) Let a(s), s ∈ [s0 , ∞) be a function such that a(s) → −∞ as s → ∞. Then there exists an infinitely differentiable, convex function k(s) such that: k1)

k(s) ≥ a(s);

k2)

k(s) ↓ −∞ as s → ∞;

k3)

k (n) (s) → 0 for all n = 1, 2, . . . .

Proof. Set a∗ (s) := sup{a(t) : t ≥ s}. Then a∗ (s) ↓ −∞ as s → ∞. Set b0 := −a∗ (s0 ) and denote as s(b), b ∈ [b0 , +∞) the function inverse to the function −a∗ (s). Let us construct a convex function that majorates s(b) and tends to infinity monotonically with all its derivatives. It can be done in the following way.

4.2. Subharmonic function with prescribed limit set

125

First we construct a piecewise linear convex function. Set s1 (b) := s0 + 1 + α0 (b − b0 ), b ∈ [b0 , b0 + 1], and choose α0 such that the inequality s1 (b) > s(b) holds for b ∈ [b0 , b0 + 1]. For this we choose α0 ≥

s(b) − s0 − 1 . b − b0 b∈[b0 ,b0 +1] sup

Since s(b) − s0 − 1 < 0 the right side is finite. For all the following intervals we set s1 (b) := s1 (b0 + j) + αj (b − b0 − j), b ∈ [b0 + j, b0 + j + 1], where αj ≥ αj−1 and satisfies the condition αj ≥

s1 (b) − s1 (b0 + j) . b − b0 − j b∈[b0 +j,b0 +j+1] sup

To obtain a smooth function, set s2 (b) :=

 α(b − x)s1 (x)dx,

(4.2.3.3)

where α(x) is defined by (2.3.1.1). Then s2 (b) is infinitely differentiable, monotonic and convex. Exercise 4.2.3.1 Check this. Set

k(s) := −s−1 2 (s),

(4.2.3.4)

s−1 2 (s)

is the inverse function to s2 . One can check that k(s) satisfies the where properties k1), k2), k3).  Exercise 4.2.3.2 Check that k(s) satisfies k1), k2), k3). Proof of Theorem 4.2.3.1. We are going to show that Φ can be taken in the form 2

Φ(x) := cek(log |x| ) |x|ρ(|x|)

(4.2.3.5)

where c and k(s) will be chosen later. Note that Φ(x) = Φ(|x|) depends only on r = |x| and pass to the variable s := log r2 . Then for φ(s) := Φ(es/2 ) we have 2 ∂ ∂ ΔΦ(x) = r1−m rm−1 cek(log r ) rρ(r) ∂r ∂r  2 m−2 ∂ ∂ = ce−s + φ(s) ≥ cme−s min[φ (s), φ (s)]. ∂s2 2 ∂s

(4.2.3.6)

126

Chapter 4. Structure of Limit Sets s

Let us chose k as in Theorem 4.2.3.2 with a(s) := log γ(r) = log γ(e 2 ). Now we estimate the derivatives from below.

 1 1 s s φ (s) = φ(s) k  (s) + sρ (e 2 ) + ρ(e 2 ) . 4 2 s

s

By k3) and k1), k  (s) → 0 and k(s) ≥ a(s). Also sρ (e 2 ) → 0 and ρ(e 2 ) → ρ by properties of proximate order (Theorem 2.8.1.4). Thus we can chose c such that 1 log γ(e 2s )+ s ρ(e s2 ) 2 e . (4.2.3.7) m Differentiating once again, we obtain

2  s s s s 1 1 1 1 φ (s) = φ(s) k  (s) + sρ (e 2 ) + ρ(e 2 ) + k  (s) + ρ (e 2 ) + sρ (e 2 ) . 4 2 2 8 φ (s) >

From here we obtain by choosing c: 1 log γ(e s2 )+ s ρ(e 2s ) 2 e . m Using (4.2.3.6), (4.2.3.7) and (4.2.3.8) we obtain: φ (s) >

s

s

(4.2.3.8)

s

ΔΦ(s) > elog γ(e 2 )+ 2 ρ(e 2 ) . Returning to the variable r we obtain (4.2.3.1). Correctness of (4.2.3.2) can be checked directly using k2) and properties of the proximate order (Theorem 2.8.1.3). 

Exercise 4.2.3.3 Check this.

4.2.4 We have already approximated distributions and subharmonic functions by infinitely differentiable functions (Theorems 2.3.4.5 and 2.6.2.3). Now we need to make more precise this approximation. Namely, we are going to make it uniform with respect to v ∈ U [ρ, σ]. We will denote ∂ l :=

∂ |l| (∂x1 )l1 (∂x2 )l2

· · · (∂xm )lm

where l = (l1 , l2 , . . . , lm ), |l| = l1 + l2 + · · · + lm . Set for v ∈ U [ρ, σ]  R v(x) := α (x − y)v(y)dy

(4.2.4.1)

(4.2.4.2)

where α is taken from (2.3.1.3). We have changed the notation from 2.3.1 and 2.6.2 because a subindex of v was already engaged for t. For a fixed 0 < δ ≤ 0.5, set Str(δ) := {x : δ ≤ |x| ≤ δ −1 }.

(4.2.4.3)

4.2. Subharmonic function with prescribed limit set

127

Theorem 4.2.4.1 (Estimation of R ) Let v ∈ U [ρ, σ]. Then R1. for a fixed g ∈ D(Rm \ 0) with supp g ⊂ Str(δ), |R v − v, g| ≤ o(1, g)2σδ −ρ

(4.2.4.4)

where o(1, g) → 0 as → 0; R2. the inequality

|∂ l R v(x)| ≤ A(m)σ −|l|−m+1 |x|−|l|+ρ ,

(4.2.4.5)

with A(m) depending only on m, holds for < |x|/2. Proof. One has R v, g = v, R g.

(4.2.4.6)

R v − v, g = v, R g − g.

(4.2.4.7)

Thus

Exercise 4.2.4.1 Check (4.2.4.6) and (4.2.4.7). 

Now |v, R g − g| ≤ max |R g − g|(x)

|v|(x)dx.

Str(δ)

(4.2.4.8)

Str(δ)

The first factor is o(1) because g is smooth. For the second one we have   |v|(x)dx ≤ 2 v + (x)dx ≤ 2σδ −ρ . (4.2.4.9) Str(δ)

Str(δ)

This and (4.2.4.8) imply R1). Differentiating the equality  R v(x) := Cm

−m α(|x − y|/ )v(y)dy,

we have



|∂ l R v(x)| ≤ Cm −|l|−m max |∂ l α(|y|)| {|y|<}

|v(x − y)|dy.

{|y|<}

Suppose |x| = 1. Then for 0 < ≤ 0.5, we have   |v|(x − y)dy ≤ |v|(x) ≤ 2 |y|<

1−<|x|<1+



1−<|x|<1+

≤ σm 2 · 2 σ(1 + ) ≤ σm 6σ ρ

v + (x)dx

128

Chapter 4. Structure of Limit Sets

where σm is the square of the unit sphere. Hence for |x| = 1 |∂ l R v(x)| ≤ A(m)σ −|l|−m+1

(4.2.4.10)

with A(m) = 6σm max |∂ l α(|y|)|. m y∈R

Set t = |x|. Apply the inequality (4.2.4.10) to v := v[t] (y) with y := x/|x|. Then |∂ l R v[t] (y)| ≤ A(m)σ −|l|−m+1 . Computing the derivatives, we obtain ∂ l R v[t] (x) = t−ρ t|l| ∂ l R v(x)|x=ty . 

Thus one has R2.

4.2.5 In this section we describe the main part of a construction that will be used in the proof of Theorem 4.2.1.2. Let {vj ∈ U [ρ, σ], j = 1, 2, . . . } and {ψj , j = 1, 2 . . . } be the partition of unity from Theorem 4.2.2.1. Let us chose j ↓ 0 such that the condition →∞ γj −m j

(4.2.5.1)

holds for γj taken from Theorem 4.2.2.1, (prtu 6). Set v(x|t) :=

∞ 

ψj (t)(vj )[t] (x),

(4.2.5.2)

j=0

where (·)[t] is defined by (3.1.2.4a). One can see that v(x|t) ∈ U [ρ, 3σ] for all t. Exercise 4.2.5.1 Show this, using properties of {ψj } and invariance of U [ρ, σ] with respect to (·)[t] . We can consider v(x|t) as a curve (a pseudo-trajectory) in U [ρ, 3σ]. Set ∞  u(x) := ψj (|x|)Rj (vj )(x)|x|ρ(|x|)−ρ . (4.2.5.3) j=0

where R is defined by (4.2.4.2). It is an infinitely differentiable function in Rm . Theorem 4.2.5.1 (Construction) One has

in D (Rm ), and

ut − v(•|t) → 0

(4.2.5.4)

Δu(x) = f (x) + γ(x)|x|ρ(|x|)−2

(4.2.5.5)

with f (x) ≥ 0 and γ(x) = o(1) as |x| → ∞.

4.2. Subharmonic function with prescribed limit set

129

Let us note that the function u(x) is “almost-subharmonic” and can be made subharmonic by summing with the function Φ from Theorem 4.2.3.1. Exercise 4.2.5.2 Prove this. So we have Theorem 4.2.5.2 (Pseudo-Trajectory Asymptotics) For any v(x|t) of the form (4.2.5.2) there exists an infinitely differentiable function u ∈ SH(ρ(r)) that satisfies (4.2.5.4). Proof of Theorem 4.2.5.1. One has ut (x) :=

∞ 

ψj (t|x|)(Rj (vj ))[t] (x)a(x, t),

j=0

where

|tx|ρ(t|x|)−ρ . tρ(t)−ρ For any 0 < δ < 0.5 and x ∈ Str(δ), a(x, t) → 1 uniformly in |x|as t → ∞. This follows from Theorem 2.8.1.3, ppo3). a(x, t) :=

Exercise 4.2.5.3 Check this in detail. We have ut (x) − v(x|t) =

∞ 

[ψj (t|x|)(Rj (vj ))[t] (x)a(x, t) − ψj (t)(vj )[t] (x)],

(4.2.5.6)

j=0

and there are no more than three summands in the sum for sufficiently large t = t(δ) because of Theorem 4.2.2.1, prtu4. Let us estimate every summand. One has bj (x, t) := [ψj (t|x|)(Rj (vj ))[t] (x)a(x, t) − ψj (t)(vj )[t] (x)] = [ψj (t|x|) − ψj (t)](Rj (vj )))[t] a(x, t) + ψj (t)(Rj (vj ))[t] (x)[a(x, t) − 1] + ψj (t)[(Rj (vj ))[t] (x) − (vj )[t] (x)] := (a1 + a2 + a3 )(x, t). Let us estimate bj (•, t), g for every g ∈ D(Rm \ 0). We can assume that supp g ⊂ Str(δ). Set M (g) := max |g|(x). x∈Str(δ)

We have |a1 (•, t), g| ≤ M (g) max |rψj (r)|δ −1

 |(Rj (vj ))[t] |(x)dx.

r∈(0,∞)

Str(δ)

130

Chapter 4. Structure of Limit Sets



One can check that

|(Rj (vj ))[t] |(x)dx ≤ 3σδ −ρ .

Str(δ)

Exercise 4.2.5.4 Check this using (4.2.4.9) and the invariance of U [ρ, σ] with respect to (•)[t] (see (3.1.2.4)). Hence |a1 (•, t), g| ≤ C1 (g)γj .

(4.2.5.7)

Let us estimate a2 (x, t). We have a2 (•, t), g ≤ max |a(x, t) − 1|ψj (t)M (g)3σδ −ρ = C2 (g)o(1)

(4.2.5.8)

Str(δ)

where o(1) → 0 as t → ∞. For estimating a3 (x, t), we use Theorem 4.2.4.1 (Estimation of R ), (4.2.4.4): |a3 (•, t), g| ≤ o( j , g)2σδ −ρ

(4.2.5.9)

where o( j , g) → 0 as j → ∞. Hence (4.2.5.7), (4.2.5.8) and (4.5.5.9) imply bj (•, t), g → 0

(4.2.5.10)

as t → ∞ and j → ∞. Suppose, for a large fixed t, the sum (4.2.5.6) contains bj (x, t) for j = j(t), j = j(t) + 1 and j = j(t) + 2. This implies that j(t) → ∞ as t → ∞. Since ut (•) − v(•|t), g = bj(t) (•, t), g + bj(t)+1 (•, t), g + bj(t)+2 (•, t), g we obtain from (4.2.5.10) that ut (•) − v(•|t), g → 0 as t → ∞ for any g ∈ D(Rm \ 0). This is (4.2.5.4). Let us prove (4.2.5.5). We have Δu =

∞ 

[Δ(Rj vj )(x)ψj (x)|x|ρ(|x|)−ρ +

j=0



∂ l (Rj vj )(x)∂ n ψj (x)∂ k |x|(ρ|x|)−ρ ],

l,m.k

(4.2.5.11) where l, m, k are multi-indexes that satisfy the condition: in any summand there are derivatives in the same variable, the derivatives of ψj and |x|ρ(|x|)−ρ have no more than second order and the derivatives of Rj vj (x) have no more than first order. Exercise 4.2.5.5 Check this.

4.2. Subharmonic function with prescribed limit set

131

As usual, the derivative of zero order is the function itself. For any x ∈ Str(δ), the outside sum contains no more then three summands. First we consider only the terms in the square brackets. The first term is nonnegative because of subharmonicity of Rj vj and non-negativity of all the other factors. Set  f (x) := [Δ(Rj vj )(x)ψj (x)|x|ρ(|x|)−ρ ≥ 0. (4.5.2.12) j=0∞

Using Theorem 4.2.4.1, R2) we obtain −|l|−m+1

|∂ l (Rj vj )(x)| ≤ A(m)σ j

|x|−|l|+ρ

(4.2.5.13)

for |l| = 0 or |l| = 1. From Theorem 4.2.2.1, prtu6), and inequality |∂xi |x|| ≤ 1 we obtain |∂ n ψj (|x|) ≤ |ψ (n) (r)||r=|x| ≤ γj |x|−|n|

(4.2.5.14)

for |n| = 1, 2. Using properties of the smooth proximate order (Theorem 2.8.1.4), one can obtain (4.2.5.15) |∂ |k| |x|ρ(|x|)−ρ | = (|x|ρ(|x|)−ρ−|k| )|r=|x| (1 + o(1)), as |x| → ∞. Exercise 4.2.5.6 Check in detail (4.2.5.13), (4.2.5.14) and (4.2.5.15). Thus, for every term of the inner sum, we have |∂ l (Rj vj )(x)∂ n ψj (x)∂ k |x|ρ(|x|)−ρ | −|l|−m+1

≤ A(m)σγj j

|x|−2+ρ |x|ρ(|x|)−ρ

≤ βj |x|ρ(|x|)−2 ,

(4.2.5.16)

where βj → 0 because of the condition (4.2.5.1). Recall that for every large x the outside sum contains no more than three summands, say, j = j(x), j = j(x) + 1 and j = j(x) + 2. Thus j(x) → ∞ as |x| → ∞. Hence (4.2.5.12) and (4.2.5.16) imply (4.2.5.5).  4.2.6 Proof of Theorem 4.2.1.2. Let v(•|t) have the form (4.2.5.2). We denote as Ω(v) a set of the D -limits of the form w := lim v(•|tk ). tk →∞

We are going to construct some v(•|t) for which Ω(v) = U,

(4.2.6.1)

and at the next step to use Theorem 4.2.5.2 to obtain a subharmonic function with the same limit set.

132

Chapter 4. Structure of Limit Sets

First we describe the construction of the function v(•|t). Let {rk , tk , k = 1, 2, . . . } be an alternating sequence r0 = 1, rk < tk < rk+1 such that lim

k→∞

tk rk+1 = lim = ∞. k→∞ tk rk

Let us chose in U a countable, dense set {gj } and form from it a sequence {wk } such that every element gj is repeated infinitely often. For example, w1 := g1 , w2 := g1 , w3 := g2 , w4 := g1 , w5 := g2 , w6 := g3 , . . . . Set qk := (wk )[1/tk ] = T− log tk wk in the notation (4.1.5.1). Now we use that (T• , U ) is chain recurrent. Set αk := log

rk ; tk

ωk := log

rk+1 tk

and find, by Theorem 4.1.2.1, a sequence {vj } ⊃ {qk } such that the condition (4.1.2.1) holds, i.e., (4.2.6.2) Tωk vk − Tαk+1 vk+1 → 0 as k → ∞. Set in Theorem 4.1.2.3 pk := Tαk+1 vk+1 , qk := Tωk vk and find γk such that the condition Tτ ◦ Tωk vk − Tτ ◦ Tαk+1 vk+1 → 0

(4.2.6.3)

holds uniformly for τ ∈ [−γk+1 , γk ]. Set

1 

tk σk := min eγk , . rk

These σk satisfy the conditions (4.2.2.1) and (4.2.2.2). Exercise 4.2.6.1 Check this. We define v(•|t) by (4.2.5.2) with described vj and with ψj from Theorem 4.2.2.1, corresponding to the chosen rj and σj . Let us prove (4.2.6.1). Consider for fixed k the following three cases. 1. t ∈ [rk σk , rk+1 /σk+1 ); 2. t ∈ [rk+1 /σk+1 , rk+1 ); 3. t ∈ [rk , rk σk ).

4.2. Subharmonic function with prescribed limit set

133

For the first case we have v(•|t) = (vk )[t/tk ] = Tlog(t/tk ) vk . For the second one v(•|t) = ψk (t)(vk )[t/tk ] + ψk+1 (t)(vk+1 )[t/tk+1 ] = (vk )[t/tk ] + ψk+1 (t)[(vk+1 )[t/tk+1 ] − (vk )[t/tk ] . We transform the expression in the square brackets (vk+1 )[t/tk+1 ] = Tlog(t/tk+1 ) vk+1 = Tlog(t/rk+1 ) ◦ Tlog(rk+1 /tk+1 ) vk+1 = Tlog(t/rk+1 ) ◦ Tαk+1 vk+1 . For the second term, we obtain (vk )[t/tk ] = Tlog(t/rk+1 ) ◦ Tωk vk . Exercise 4.2.6.2 Check this. Setting τ := log(t/rk+1 ), we have v(•|t) = (vk )[t/tk ] + ψk+1 (t)[Tτ ◦ Tαk+1 vk+1 − Tτ ◦ Tωk vk ],

(4.2.6.4)

where τ ∈ [− log σk+1 , 0) ⊂ [−γk+1 , γk ]. For the third case, set τ := log(t/rk ). Then v(•|t) = (vk )[t/tk ] + ψk (t)[Tτ ◦ Tωk−1 vk−1 − Tτ ◦ Tαk vk ],

(4.2.6.5)

where τ ∈ [0, log σk ) ⊂ [−γk , γk−1 ]. Let tN → ∞ be an arbitrary sequence. Choosing a subsequence, we may suppose that there exist the limits (vk(tN ) )[tN /tk(tN ) ] → v ∗ ∈ U and v(•|tN ) → v∞ . Choosing a subsequence, we may suppose that tN satisfies either 1 or 2 or 3 For case 1, we obtain at once v∞ = v ∗ ∈ U . For case 2, from (4.2.6.4), (4.2.6.2) and Theorem 4.1.1.3 we obtain that the superfluous addends tend to zero, and hence v∞ ∈ U . The same holds for case 3. Hence Ω(v) ⊂ U . Further, for t = tk , we have v(•|t) = wk . The sequence {wk } contains the set {gj } that is dense in U. Thus Ω(v) ⊃ U . Thus equality (4.2.6.1) has been proved. As already said, the application of Theorem 4.2.5.2 concludes the proof. 

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Chapter 4. Structure of Limit Sets

4.3 Further properties of limit sets 4.3.1 Let as mark the following property of the pseudo-trajectory v(•|t) defined in (4.2.5.2): Theorem 4.3.1.1 One has Tτ v(•|et ) − v(•|et+τ ) → 0

(4.3.1.1)

as t → ∞ uniformly with respect to τ ∈ [a, b] for any [a, b] ⊂ (−∞, ∞). Proof. Using the definition of (•)t (see (3.1.2.1)) and (4.2.5.4), the remainder in (4.3.1.1) can be represented in the form b(t, τ, •) := Tτ v(•|et ) − v(•|et+τ ) = Tτ (uet ) − uet+τ + o(1) where o(1) → 0 uniformly with respect to τ ∈ [a, b] for any [a, b] ⊂ (−∞, ∞). Exercise 4.3.1.1 Check this in detail. Then we obtain t

b(t, τ, •) = uet+τ [eρ(e

)−ρ(et+τ )

− 1] + o(1) → 0

uniformly in the same sense due to precompactness of the family {uet } and properties of the proximate order. Exercise 4.3.1.2 Check this in detail.



The property (4.3.1.1) shows that the pseudo-trajectory v(•|t) behaves asymptotically like the dynamical system T• . Thus a pseudo-trajectory with this property is called an asymptotically dynamical pseudo-trajectory with dynamical asymptotics T• (a.d.p.t.). Theorem 4.2.5.1 shows that for any a.d.p.t. of the form (4.2.5.2) there exists u ∈ SH(ρ(r)) that satisfies the condition uet − v(•|et ) → 0

(4.3.1.2)

as t → ∞. The following assertion shows that we can suppose v(•|•) to be an arbitrary, in some sense, a.d.p.t. We call a pseudo-trajectory w(•|•) piecewise continuous if the property w(•|t + h) − w(•|t) → 0

(4.3.1.3)

as h → 0 holds for all t except perhaps a countable set without points of condensation.

4.3. Further properties of limit sets

135

Let U ⊂ U [ρ, σ] for some σ > 0. A pseudo-trajectory w(•|•) is called ω-dense in U if Ω(w) = U (see (4.1.3.2)), i.e., {v ∈ U [ρ] : (∃tj → ∞) v = D − lim w(•|etj )} = U.

(4.3.1.4)

We have proved already that v(•|•) defined by (4.2.5.2) has this property (see (4.2.6.1)). Now we consider again the dynamical system (T• , U ) where U ⊂ U [ρ, σ] for some σ > 0 and Tt is defined by (4.2.1.1). Theorem 4.3.1.2 (A.D.P.T. and Chain Recurrence) (T• , U ) is chain recurrent iff there exists an a.d.p.t. that is piecewise continuous and ω-dense in U . Necessity has been proved already, because the pseudo-trajectory (4.2.6.2) possesses this property. Sufficiency will be proved later. The claim of piecewise continuity can be justified by Theorem 4.3.1.3 For any u ∈ SH(ρ(r)) there exists a piecewise continuous pseudotrajectory w(•|•) such that (4.3.1.5) ut − w(•|t) → 0 as t → ∞. Of course, w(•|•) is a.d.p.t. Exercise 4.3.1.2 Check this. 4.3.2 Proof of Theorem 4.3.1.3. Let {tn } be any sequence such that tn → ∞, tn+1 /tn → 1,

(4.3.2.1)

for example, tn = n. There exists a sequence {vn } ⊂ Fr[u] such that utn − vn → 0.

(4.3.2.2)

w(•|t) := vn , for tn < t ≤ tn+1 .

(4.3.2.3)

Set This is a piecewise continuous function. Let us prove that ut − w(•|t) → 0. Assume the opposite; i.e., there exists a sequence can suppose that

(4.3.2.4) {tk }

such that it is not true. We

utk → w1 ∈ Fr[u], w(•|tk ) → w2 ∈ Fr[u], w1 = w2 .

(4.3.2.5)

136

Chapter 4. Structure of Limit Sets

Let us find a sequence {nk } such that tnk < tk < tnk +1 . Then tnk /tk → 1.

(4.3.2.6)

From (4.3.2.5), (4.3.2.3) and (4.3.2.2) we have utnk → w2 .

(4.3.2.7)

Then we have, using properties of (•)t and the proximate order, utk = (utnk )[tk /tnk ] (1 + o(log(tk /tnk )) → w2 because of (4.3.2.6) and the continuity of u[t] on (u, t). However (4.3.2.8) contradicts (4.3.2.5). Thus (4.3.2.4) holds.

(4.3.2.8) 

4.3.3 Now we will prepare the proof of Theorem 4.3.1.2. Let {vk , k = 1, 2, . . . } ⊂ U [ρ, σ] for some σ be a sequence of functions and {rk , k = 1, 2, . . . }, {tk k = 1, 2, . . . } be two sequences such that 0 < rk < tk < rk+1 , k = 1, 2, . . .

(4.3.3.1)

lim tk /rk = lim rk+1 /tk = ∞.

(4.3.3.2)

w∗ (•|t) := (vk )t/tk , for t ∈ [rk , rk+1 )

(4.3.3.3)

and k→∞

Set

k→∞

where k = 1, 2, . . . . Theorem 4.3.3.1 Let w(•|•) ⊂ U be an arbitrary ω-dense a.d.p.t. and {pj , j = 1, 2, . . . } ⊂ U an arbitrary sequence. Then there exists a sequence {vk , k = 1, 2, . . . }⊃ {pj , j = 1, 2, . . . } and sequences {rk , k = 1, 2, . . . } and {tk , k = 1, 2, . . . } satisfying (4.3.3.1) and (4.3.3.2) such that for w∗ (•|•) determined by (4.3.3.3) the condition (4.3.3.4) w∗ (•|t) − w(•|t) → 0 as t → ∞ is fulfilled. This proposition shows that any ω-dense a.d.p.t. is equivalent to one constructed of long pieces of trajectories of the dynamical system T• . Proof of Theorem 4.3.3.1. We can take sequences { j ↓ 0, j = 1, 2, . . . } and {bj ↑ ∞, j = 1, 2, . . . } and choose a sequence {τj , j = 1, 2, . . . } such that the inequalities (4.3.3.5) d(Tτ pj − Tτ w(•|τj )) < j /2 and d(Tτ w(•|t) − w(•|eτ t)) < j /2, t > τj are fulfilled uniformly with respect to τ ∈

2 [b−2 j+1 , bj ].

(4.3.3.6)

4.3. Further properties of limit sets

137

Indeed, w(•|•) is ω-dense in U , and hence we can find τn → ∞ such that pn − w(•|τn ) → 0. Set in Lemma 4.1.2.2, pn := pn , qn := w(•|τn ), γn := 2 log bj . Then for any j we can find τj := τnj such that (4.3.3.5) holds uniformly with 2 respect to τ ∈ [b−2 j+1 , bj ]. The inequality (4.3.3.6) holds, because w(•|•) is asymptotically dynamical (see (4.3.1.1)). We can also suppose without loss of generality that τj > τj−1 b2j−1 , i.e., that the sequence {τj } is rather thin. The inequality (4.3.3.5) shows that for intervals of t that are determined by 2 the inequality b−2 j+1 ≤ t/τj ≤ bj our pseudo-trajectory is already close to some trajectories. Now we divide the spaces between such intervals into equal parts in the logarithmic scale such that their logarithmic lengths would be between log bj and log bj+1 , so that they tend to infinity. To this end, set

 log τj+1 − log τj nj := bj where [·] means the entire part, and 1

γj := (τj+1 /τj ) 2nj . It is clear that bj < γj < b2j . As centers of new intervals we take the points τj,l := τj γj2l , l = 0, 1, . . . , nj . Thus τj,0 = τj and τj,nj = τj+1 . The ends of the intervals are τj,l /γj and τj,l γj . Now we complete the sequence {pj } by the values of the pseudo-trajectory w(•|t) in the centers of the intervals, i.e., we set pj,l := w(•|τj,l ), l = 1, . . . , nj − 1 For t ∈ (τj,l /γj , τj,l γj ), l = 1, . . . , nj − 1 we have d((pj,l )t/τj,l − w(•|t)) < j /2

(4.3.3.7)

because of (4.3.3.6). For l = 0 and l = nj we set accordingly pj,0 := pj ; pj,nj := pj+1 . Using (4.3.3.5) and (4.3.3.6) we have an inequality like (4.3.3.7) for l = 0, l = nj but with j instead of j /2. We complete the proof, re-denoting all the centers τj,l as tk , all the ends as  rk and all the pj,l as vk .

138

Chapter 4. Structure of Limit Sets

4.3.4 Proof of sufficiency in Theorem 4.3.1.2. A direct corollary of the previous Theorem 4.3.3.1 is (4.3.4.1) w∗ (•|rk − 0) − w∗ (•|rk ) → 0 as k → ∞. Really, w∗ (•|•) is an a.d.p.t. Exercise 4.3.4.1 Check this as in Theorem 4.3.1.1 using that w(•|•) is asymptotically dynamical. For τ ∈ [− , 0] and t = rk we have uniformly on τ , Tτ w∗ (•|t) − w∗ (•|t) = Tτ (vk )rk /tk − (vk+1 )rk /tk → 0. Setting τ = 0 we obtain (4.3.4.1). Let V ⊂ U be an arbitrary open set, > 0 arbitrarily small and s > 0 arbitrarily large. We show that there exists an ( , s)-chain from V to V . Choose s1 such that i. for rk > s1 , d(w∗ (•|rk − 0), w∗ (•|rk )) < . This is possible by virtue of (4.3.4.1). ii. w(•|s1 ) ∈ V . This is possible because w(•|•) is ω -dense. iii. d(w∗ (•|t), w(•|t)) < d(w(•|s1 ), ∂V ) for t > s1 . This is possible because of Theorem 4.3.3.1. Choose s2 > s1 such that w(•|s2 ) ∈ V . This is possible because w(•|•) is ω-dense. Then the pseudo-trajectory w∗ (•|et ) for s1 ≤ et ≤ s2 is an ( , s)-chain connecting w∗ (•|s1 ) and w∗ (•|s2 ) that belong to V . Exercise 4.3.4.2 Check this in detail. Hence (T• , U ) is chain recurrent. 4.3.5 We will prove one more existence theorem that is a corollary of Theorem 4.2.1.2. Theorem 4.3.5.1 Let Λ ⊂ U [ρ] be a compact connected and T• -invariant subset of U [ρ]. Then for any proximate order ρ(r) → ρ there exists u ∈ SH(Rm , ρ, ρ(r)) such that h(x, u) = sup{v(x) : v ∈ Λ}, h(x, u) = inf{v(x) : v ∈ Λ}.

(4.3.5.1) (4.3.5.2)

Proof. Let U := ConvΛ be the convex hull of Λ. It is linearly connected and hence polygonally connected (see 4.1.4). By Theorem 4.1.4.1 it is chain recurrent and by Theorem 4.2.1.2 for any proximate order ρ(r) → ρ there exists u ∈ SH(Rm , ρ, ρ(r))

4.3. Further properties of limit sets

139

such that Fr[u, ρ(r), Vt , Rm ] = U. Since every v ∈ U can be represented in the form v = av1 + (1 − a)v2 for 0 ≤ a ≤ 1, v1 , v2 ∈ Λ we obtain (4.3.5.1) and (4.3.5.2) from Theorem 3.2.1.1 (Properties of Indicators), h2). Exercise 4.3.5.1 Check this. 4.3.6 In applications we need the following Theorem 4.3.6.1 Let p ∈ P ⊂ Rm and let P be a connected closed set. Let UP := {v(z, p) : p ∈ P ⊂ Rm } be a family of functions with parameter p such that for every p ∈ P , v(•, p) ∈ U [ρ] and satisfy the condition (4.1.3.3). Then there exists u ∈ SH(ρ(r)) such that Fr[u] = UP . This is a direct corollary of Theorems 4.1.1.2, 4.1.3.3 and 4.2.1.2. Exercise 4.3.6.1 Explain this in detail. 4.3.7 In the next three sections we return to the periodic limit sets (see Theorem 4.1.7.1). We show that the limit set Fr[u, ρ(r), V• , Rm ] of every subharmonic function u ∈ SH(ρ(r), Rm ), ρ(r) → ρ for non-integer ρ can be approximated in some sense by periodic limit sets ([Gi], [GLO, Ch. 3, § 2, Thm. 10]). Here we give some definitions. Let Xn ⊂ U [ρ], n = 1, 2, . . . be a sequence of compact sets. We say that Xn converges to a compact set Y ⊂ U [ρ], i.e., D − lim Xn = Y n→∞

(4.3.7.1)

if the following two conditions hold: converg1)

∀xn ∈ Xn , n = 1, 2, . . . ∃xnj ∈ Xnj , j = 1, 2, . . . and y ∈ Y such that D − lim xnj = y;

converg2)

∀y ∈ Y ∃xn ∈ Xn , n = 1, 2, . . . , such that xn → y.

j→∞

On every compact set K in D -topology one can introduce a metric d(•, •) such that the topology generated by this metric is equivalent to D -topology (see, e.g., [AG(1982)]). Denote by X := {y ∈ K : ∃x ∈ X such that d(y, x) < } the -neighborhood of X. Let X, Y be two compact sets. Set d(X, Y ) := inf{ : X ⊂ Y , Y ⊂ X }.

140

Chapter 4. Structure of Limit Sets

Exercise 4.3.7.1 Prove the assertion (4.3.7.1) ⇐⇒ {d(Xn , Y ) → 0}.

(4.3.7.2)

We prove the following Theorem 4.3.7.1 (Approximation by Periodic Limit Sets) Let u ∈ SH(ρ(r), Rm ), ρ(r) → ρ for non-integer ρ. Then for every V• there exists a sequence un ∈ SH(ρ(r), Rm ) with periodic limit sets Fr[un , ρ(r), V• , Rm ] such that Fr[un , •] → Fr[u, •]. This theorem is a corollary of the following Theorem 4.3.7.2 Let μ ∈ M(ρ(r), Rm ), ρ(r) → ρ for non-integer ρ. Then for every V• there exists a sequence μn ∈ SH(ρ(r), Rm ) with periodic limit sets Fr[μn , ρ(r), V• , Rm ] such that Fr[μn , •] → Fr[μ, •]. Proof of Theorem 4.3.7.1. The canonical potential u(x) := Π(x,μ,p) (see (2.9.2.1)) of a measure μ ∈ M(ρ(r), Rm ) belongs to SH(ρ(r), Rm ) by Theorem 2.9.3.3 and has a limit set Fr[u, •] = {Π(•, ν, p) : ν ∈ Fr[μu ]} by Theorem 3.1.4.4 (*Hadamard). The potentials un (x) := Π(x, μn , p) have periodic limit sets Fr[un , •] = {Π(•, ν, p) : ν ∈ Fr[μun ]} by Theorem 3.1.5.0. Let us prove that Fr[un , •] =: Xn → Y := Fr[u, •]. If vn ∈ Fr[un , •] then from the corresponding sequence of νn := νvn ∈ Fr[μn , •] we can find a subsequence νnj and ν ∈ Fr[μ, •] such that νn → ν (by Theorem 2.2.3.2 (Helly)). It is easy to check, using Theorem 3.1.4.3 (*Liouville), that v ∗ = D − lim Π(•, νnj ) exists and coincides with v = Π(•, ν, p) ∈ Fr[u, •]. j→∞

So the condition converg1) is verified. In the same way one can check converg2).  Exercise 4.3.7.2 Prove this in detail. 4.3.8 Now we are going to prove Theorem 4.3.7.2. We begin from Proposition 4.3.8.1 For any μ ∈ M(ρ(r), •) there exists μ ˆ ∈ M(ρ, •) such that Fr[ˆ μ, ρ, •] = Fr[μ, ρ(r), •]. In other words we can suppose further that ρ(r) ≡ ρ.

(4.3.8.0)

4.3. Further properties of limit sets

141

Proof. Set L(r) = rρ(r)−ρ and μ ˆ(dx) := L−1 (|x|)μ(dx).

(4.3.8.1)

Using properties of proximate order (Section 2.8.1., po1)–po4)), it is easy to check that [L−1 (r)] = L−1 (r)o(1) and [L(r)] = L(r)o(1), as → 0.

(4.3.8.2)

Exercise 4.3.8.1 Prove this. Let us show that μ ˆ ∈ M[ρ, Δ] for some Δ. Indeed μ ˆ (R) = R−ρ−m+2 ρ+m−2 R

R

μ(r) μ(dr) = ρ+m−2 |R + R−ρ−m+2 L(r) R L(r) 0

0

R

μ(r)(L−1 ) dr.

0

We suppose that μ(r) = 0 in some neighborhood of zero. Using (4.3.8.2) we obtain further for the last expression, μ(R)R

−ρ(r)

+R

−ρ−m+2

R

μ(r)(L−1 )o(1)dr.

0

Using the l’Hospital rule, we obtain lim R

−ρ−m+2

R

R→∞

μ(r)(L−1 (r))o(1/r)dr = (−ρ − m + 2) lim μ(R)R−ρ(R) o(1/R). R→∞

0

Thus lim sup R→∞

μ ˆ (R) ≤ lim sup μ(R)(L−1 (R))[1 + o(1/R)] = Δ[μ, ρ(r)] < ∞. ρ+m−2 R R→∞

μ[t] . This implies equality (4.3.8.0) because L(t) → 1 as Let us note that μt = L(t)ˆ t → ∞. Exercise 4.3.8.2. Prove this in detail.



Proof of Theorem 4.3.7.2. As we already said we can also suppose that μ ∈ M[ρ]. Let ν ∈ Fr[μ]. We can suppose that ν({|x| = 1}) = 0.

(4.3.8.2a)

Otherwise we can find τ such that ν[τ ] ({|x| = 1}) = 0 and if νn → ντ and are periodic, then (νn )[1/τ ] are also periodic and (νn )[1/τ ] → ν.

142

Chapter 4. Structure of Limit Sets

Let rn → ∞ be such that μ[rn ] → ν. By passing to subsequences we can make rn+1 /rn > rn . Denote Kn := {x : rn ≤ |x| < rn+1 }. Set for every E ⊂ Kn μn |E := μ(E) and continue it periodically with the period Pn = rn+1 /rn by the equality μn (Pnk E) = Pnkρ μ(E), k = ±1, ±2, . . . . Since every X ∈ R

m

(4.3.8.3)

can be represented in the form X=

∞ 

{X ∩ Pnk Kn },

k=−∞

we can define μn (X) :=

∞ 

μn ({X ∩ Pkn Kn }).

k=−∞

It is easy to check that μn is periodic with period Pn and μn ∈ M[ρ, Δ] with Δ independent of n. Exercise 4.3.8.3 Check this. Let us prove that Fr[μ] = lim Fr[μn ].

(4.3.8.4)

n→∞

Check the condition converg1). Let νnj ∈ Fr[μnj ] and suppose D − lim νnj := ν. j→∞

Let us prove that ν ∈ Fr[μ]. Since Fr[μnj ] is a periodic limit set, νnj = (μnj )[τj ] . Take kj such that

τj := τj Pnkjj ∈ [rnj , rnj +1 ).

From periodicity μnj we obtain νnj = (μnj )[τj ] . Passing to a subsequence if necessary, we can consider three cases: i) lim τj /rnj = ∞, lim τj /rnj +1 = 0; j→∞

j→∞

ii) lim τj /rnj = τ ; 1 ≤ τ < ∞; In this case we have also lim τj /rnj +1 = 0. j→∞

j→∞

iii) lim τj /rnj +1 = τ ; 0 < τ ≤ 1; In this case we have also lim τj /rnj = ∞. j→∞

j→∞

Consider the case i). Let φ ∈ D(Rm \ O). Then supp φ(x/τj ) ⊂ (rnj , rnj +1 ) for j ≥ j0 . It is easy to see that, for j ≥ j0 , (μnj )[τj ] , φ = μ[τj ] , φ.

4.3. Further properties of limit sets

143

Exercise 4.3.8.4 Check this. Since μ[τj ] → ν ∈ Fr[μ] by definition the condition converg1) holds for the case i). Consider the case ii). Recall that O ∈ / supp φ. Then there exists 1 ≤ c < ∞ such that supp φ ⊂ {x : |x| ∈ (1/c, c)}. Define φt (x) := φ(x/t)(1/t)ρ . Represent τj in the form τj := ej τ rnj

where

ej :=

τj rnj τ

.

The condition ii) means that ej → 1.

(4.3.8.4a)

Compute νj , φ := (μnj )[τj ] , φ = μnj , ((φτ )ej )rnj . Note that supp φτ ⊂ {x : |x| ∈ (τ /c, τ c)}. We can increase c so that 1 ∈ (τ /c, τ c). Consider the following partition of unity. Choose the functions ηk ∈ D(Rm ), k = 1, 2, 3 so that η1 (t) + η2 (t) + η3 (t) = 1 for t ≥ 1 and supp η1 ⊂ {x : |x| < 1 − }, supp η2 ⊂ {x : |x| ∈ (1 − 2 , 1 + 2 )}, supp η3 ⊂ {x : |x| > 1 + }, where is an arbitrary number, satisfying τ /c < 1 − 2 < 1 + 2 < τ c. Represent φτ in the form φτ = ψ1 + ψ2 + ψ3 , In this notation νj , φ =

where 3  k=1

ψk = φτ ηk , k = 1, 2, 3.

μnj , (ψk )ej rnj .

(4.3.8.5)

144

Chapter 4. Structure of Limit Sets

Choose j such that for j ≥ j the following inclusions hold: supp(ψ1 )ej rnj ⊂ {x : |x| ∈ (rnj τ /c, rnj (1 − ))}; supp(ψ2 )ej rnj ⊂ {x : |x| ∈ ((1 − )rnj , rnj (1 + ))}; supp(ψ3 )ej rnj ⊂ {x : |x| ∈ ((1 + )rnj , τ rnj }. Thus for ψ3 we have  μnj , (ψ3 )ej rnj  = = D



(ej rnj )−ρ ψ3 (|x|/(ej rnj ))μnj (dx) (ej rnj )−ρ ψ3 (|x|/(ej rnj ))μ(dx) = μ[rnj ] , (ψ3 )ej . D

Since μ[rnj ] →ν and (ψ3 )ej → ψ3 we have (see Theorem 2.3.4.6) lim μnj , (ψ3 )ej rnj  = ν, ψ3 .

j→∞

(4.3.8.6)

Consider the addend with ψ1 . Because of periodicity μnj we have μnj , (ψ1 )ej rnj  = (μnj )[Pnj ] , (ψ1 )ej rnj . Transforming the RHS we obtain (μnj )[Pnj ] , (ψ1 )ej rnj  = μnj , ((ψ1 )Pnj )ej rnj . Since Pnj = rnj +1 /rnj the following inclusion holds for j ≥ j : τ supp(ψ1 )Pnj ej rnj ⊂ {x : |x| ∈ (Pnj rnj , Pnj rnj (1 − ))} c τ = {x : |x| ∈ (rnj +1 , rnj +1 (1 − ))} ⊂ {x : |x| ∈ (rnj , rnj +1 )}. c Thus μnj , ((ψ1 )Pnj )ej rnj  = μ, (ψ1 )Pnj ej rnj  = μ, (ψ1 )ej rnj  = μ[rnj ] , (ψ1 )ej . Hence lim μnj , (ψ1 )ej rnj  = ν, ψ1 

j→∞ D

because ej → 1 and μ[rnj ] → ν (see Theorem 2.3.4.6). From (4.3.8.5), (4.3.8.6) and (4.3.8.7) we obtain lim νj , φ = ν, ψ1 + ψ3  + lim μnj , ((ψ2 )ej rnj .

j→∞

j→∞

(4.3.8.7)

4.3. Further properties of limit sets

145

Let us estimate the last limit. We have lim μnj , (ψ2 )ej rnj  = (μnj )[ej rnj ] , ψ2 .

j→∞

Define E1 ( ) := {x : |x| ∈ (1 − 2 , 1)}; E2 ( ) := {x : |x| ∈ [1, 1 + 2 )}. Suppose is chosen so that ν(∂Ek ) = 0, k = 1, 2.

(4.3.8.7a)

Recall that ν satisfies the condition (4.3.8.2a), hence E1 , E2 are ν-squarable and hence (see Theorem 2.2.3.7) lim μ[rn ] (Ek ( )) = ν(Ek ( )), k = 1, 2.

n→∞

Define Cφ := max{φ(x) : x ∈ Rm }. Then for j ≥ j , |(μnj )[ej rnj ] , ψ2 | ≤ Cφ (μnj )[ej rnj ] (E1 ( ) ∪ E2 ( )) = Cφ ((μnj )[ej rnj ] (E1 ( )) + (μnj )[ej rnj ] (E2 ( )). By definition (μnj )[ej rnj ] (E2 ( )) = μ[ej rnj ] (E2 ( )). Because of (4.3.8.4a) we obtain lim μ[ej rnj ] (E2 ( )) = ν(E2 ).

j→∞

Exercise 4.3.8.5 Check in detail. To compute the limit of the first addend we use periodicity of μnj : μnj (Pnj E1 ( )) = (μnj )[Pnj ] (E1 ( )), μnj (E1 ( )) = Pn−ρ j where rnj Pnj E1 ( ) = {x : |x| ∈ (rnj +1 (1 − 2 ), rnj +1 )}. Thus (μnj )[ej rnj ] (E1 ( )) = (μnj )[ej rnj Pnj ] (E1 ( )) = (μnj )[ej rnj +1 ] (E1 ( )) = μ[ej rnj +1 ] (E1 ( )). From this we obtain lim (μnj )[ej rnj ] (E1 ( )) = ν(E1 ( )).

j→∞

146

Chapter 4. Structure of Limit Sets

Therefore lim |(μnj )[ej rnj ] , ψ2 | ≤ Cφ ν(E1 ( ) ∪ E2 ( )).

j→∞

(4.3.8.8)

Because of (4.3.8.2a) we have ν({x : |x| ∈ (1 − 2 , 1 + 2 )}) → 0

(4.3.8.9)

as → 0 over the set of satisfying (4.3.8.7a). From (4.3.8.8) and (4.3.8.9) we obtain 2 3 (4.3.8.10) lim (μnj )[ej rnj ] , ψ2 → 0 j→∞

and ν, ψ2  → 0

(4.3.8.11)

when → 0. Hence if → 0 satisfying (4.3.8.7a) we have lim νj , φ − ν[τ ] , φ

j→∞

= lim [ν, ψ1 + ψ3  + lim (μnj )[ej rnj ] , ψ2  − ν, ψ1 + ψ2 + ψ3 ] →0

j→∞

= lim [ lim (μnj )[ej rnj ] , ψ2  − ν, ψ2 ] = 0. →0 j→∞

The last equality holds because every addend tends to zero. The case iii) can be considered in an analogous way. Exercise 4.3.8.6 Consider it. Thus the condition converg1) was checked.



4.3.9 Now we should check the condition converg2). We need Lemma 4.3.9.1 Let μ ∈ M[ρ], ν ∈ Fr[μ] and rn → ∞, n = 1, 2, . . . be a sequence such that (4.3.9.1) D − lim μ[rn ] = ν0 . n→∞

Then passing if necessary to a subsequence, we can find {rn } such that for arbitrarily ν ∈ Fr[μ] a sequence tj → ∞ exists such that D − lim μ[tj ] = ν j→∞

(4.3.9.2)

and for every n we can find tj ∈ [rn , rn+1 ]. Proof. Note that if the assertion of the lemma is satisfied for the sequence {rn , n = 1, 2, . . . } it is satisfied for every subsequence of {rn , n = 1, 2, . . . }. Let M be a countable set that is dense in Fr[μ]. Since reduction D -topology on M[ρ] is metrizable, it is sufficient to prove that we can choose a subsequence rn for which assertion of the lemma is satisfied for all ν ∈ M. We can do it using a diagonal process.

4.4. Subharmonic curves. Curves with prescribed limit sets

147

Let rn0 → ∞ be an arbitrary sequence such that D − lim μ[rn0 ] = ν0 n→∞

and let a sequence

t1j

satisfy the condition D − lim μ[t1j ] = ν1 . j→∞

{rn0 }

0 the ends of the segments [rn0 , rn+1 ] that do not conOmitting in the sequence 1 tain elements {tj } we obtain a subsequence {rn }. Continuing in such a way we obtain a subsequence {rnm } satisfying (4.3.9.1) and the sequences {t1j }, {t2j }, . . . , {tm j }, j = 1, 2 . . . satisfying

D − lim μ[tlj ] = νl , l = 1, 2, . . . m. j→∞

(4.3.9.3)

Taking a diagonal sequence {rnn }, n = 1, 2, . . . we observe that it is a subsequence  of every subsequence {rnm } and hence satisfies the assertion of the lemma. Proof of converg2). We can suppose that {rn } from the construction of μn with periodic limit sets satisfies the assertion of Lemma 4.3.9.1. Let ν ∈ Fr[μ] and μtj → ν under condition tj ∈ [rn , rn+1 ]. We should consider as in the proof of converg1) three cases i), ii) and iii). But all these cases were already considered and hence it was proved that νn := (μn )rn → ν. 

Exercise 4.3.9.1 Check this.

4.4 Subharmonic curves. Curves with prescribed limit sets 4.4.1 In this paragraph we consider subharmonic functions u ∈ SH(ρ(r)) in the plane of finite type with respect to some proximity order ρ(r) → ρ. The pair u := (u1 , u2 ), u1 , u2 ∈ SH(ρ(r)) is called a subharmonic curve (which for brevity we will refer to simply as a curve). The family (u)t := ((u1 )t , (u2 )t ) is precompact in the topology of convergence in D -topology on every component. The set of all limits Fr[u] := {v = (v1 , v2 ) : ∃tj → ∞, v = D − lim utj } j→∞

is called the limit set of the curve u. Actually this set describes coordinated asymptotic behavior of pairs of subharmonic functions.

148

Chapter 4. Structure of Limit Sets

Theorem 4.4.1.1 Fr[u] is closed, connected, invariant with respect to (•)[t] (see 3.1.2.4a) and is contained in the set U [ρ, σ] := {v = (v1 , v2 ) : vn (z) ≤ σn |z|ρ , vn (0) = 0, n = 1, 2.} where σ := (σ1 , σ2 ) Exercise 4.4.1.1 Prove this by using Theorem 3.1.2.2. Let us define σ > 0 as σn > 0, n = 1, 2. Set  U [ρ] := U [ρ, σ]. σ>0

We will write U ⊂ U [ρ] if U ⊂ U [ρ, σ] for some σ. Since (T• , U [ρ]) is a dynamical system we have two theorems analogous to Theorems 4.2.1.1 and 4.2.1.2. Exercise 4.4.1.2 Formulate and prove these theorems. All the other assertions and definitions of Sections 4.2,4.3 can be repeated for subharmonic curves. Let U ⊂ U [ρ]. Set U  := {v  : ∃v  : (v  , v  ) ∈ U }. This is a projection of U . Set for v  ∈ U  , U  (v  ) := {v  : (v  , v  ) ∈ U }. This is the fibre over v  . Theorem 4.4.1.2 Let U  U [ρ] be closed and invariant and assume that every fiber U  (v  ) is convex. Let U  = Fr[u ] for some u ∈ U (ρ(r)). Then there exists u ∈ U (ρ(r)) such that Fr(u , u ) = U . We construct a pseudo -trajectory asymptotics in the form (4.2.5.2) replacing u with u and v with v. We can directly check that this curve satisfies the assertion of the theorem. Exercise 4.4.1.3 Check this. Theorem 4.4.1.3 (Concordance Theorem) Let u ∈ U (ρ(r)) and v 0 ∈ Fr[u], and suppose v ∈ U [ρ] has the property lim Tτ v = lim Tτ v = v˜.

τ →−∞

τ →+∞

4.4. Subharmonic curves. Curves with prescribed limit sets

149

Then there exists a function w ∈ U (ρ(r)) such that the limit set of the curve u = (u, w) Fr[u] = (Fr[u], C(v)) and for every sequence tn → ∞ such that lim wtn = v, n→∞

lim utn = (v 0 , v).

(4.4.1.3.)

n→∞

For proving this theorem we should use a.d.p.t. (4.2.5.2). If vj = v 0 we replace vj by v j := (v 0 , v). If vj = v 0 we replace vj by v j := (vj , v˜). Exercise 4.4.1.3 Do that and exploit Theorem 4.3.1.2 and Theorem 4.2.1.2. Corollary 4.4.1.4 Under conditions of Theorem 4.4.1.3, if lim wtn = Tτ v, then n→∞

lim utn = Tτ v 0 .

n→∞

We should apply Tτ to (4.4.1.3) and use its continuity in D -topology.



Chapter 5

Applications to Entire Functions 5.1 Growth characteristics of entire functions 5.1.1 Let f (z) be an entire function. The function u(z) := log |f (z)| is subharmonic in R2 (= C). Hence the scale of growth subharmonic functions considered in Section 2.8 is transferred completely to entire functions. We will mark passing to entire function by changing index u for index f . For example, M (r, f ) := M (r, log |f |), T (r, f ) := T (r, log |f |). If u(z) := log |f (z)| has order ρ[u] = ρ, then f (z) has order ρ[f ] := ρ and so on. We will write f ∈ A(ρ, ρ(r)) and say “f is an entire function of order ρ and normal type with respect to proximate order ρ(r)” if log |f | is a subharmonic function of order ρ and normal type with respect to the same proximate order. Shortly, if log |f | ∈ SH(ρ, ρ(r), R2 ), then f ∈ A(ρ, ρ(r)). Exercise 5.1.1.1 Give definitions of T (r, f ), M (r, f ), ρT [f ], ρM [f ], σT [f, ρ(r)], σM [f, ρ(r)] and reformulate all the assertions of Section 2.8 in terms of entire and meromorphic functions. 5.1.2 A divisor of zeros of an entire function can be represented as an integer mass distribution n on a discrete set {zj } ⊂ C. The multiplicity of a zero zj is the mass concentrated at the point zj . The notation for characteristics of the behavior of zeros will mimic that of the behavior of masses, replacing μ for n. For example, n(Kr ), n(r) is the number of zeros (with multiplicities) in the disk Kr , ρ[n] is the convergence exponent, Δ[n] is the upper density and so on.

152

Chapter 5. Applications to Entire Functions

Exercise 5.1.1.2 Give definitions of N (r, n), ρN [n], ΔN [n], p[n]. 5.1.3 The limit set Fr[f ] of an entire function f ∈ A(ρ, ρ(r)) is defined as the limit set of the subharmonic function u(z) := log |f (z)| ∈ SH(ρ, ρ(r), R2 ) (see Section 3.1), i.e., Fr[f ] := Fr[log |f |]. (5.1.3.1) It possesses, of course, all the properties described in Chapters 3, 4 but it is not clear now if there exists an entire function with prescribed limit set, i.e., whether the subharmonic function in Theorem 4.2.1.2 can be chosen to be log |f (z)| where f ∈ A(ρ, ρ(r)). It turns out that this is possible and we prove this in Section 5.3. As it was mentioned in 3.1.1 the general form of V• for the case of the plane is Vt z = zeiγ log t , where γ is real. The limit set Fr[n] of a divisor n is the limit set of the corresponding mass distribution n (see 3.1.3). Of course generally speaking nt (see (3.1.3.2)) is not an integer mass distribution. Exercise 5.1.3.1 Give a complete definition of Fr[f ] and Fr[n], and reformulate all the theorems of Sections 3.1.2, 3.1.3 in terms of entire functions and their zeros. The connection between Fr[f ] and Fr[n] is preserved completely (see Section 3.1.5). Exercise 5.1.3.2 Reformulate the theorems of Section 3.1.5 for entire functions. 5.1.4 Let f = f1 /f2 be a meromorphic function, where f1 , f2 have no common zeros. If f2 (0) = 1, f1 (0) = 0 and f1 , f2 ∈ A(ρ, ρ(r)), then u := log |f1 | − log |f2 | ∈ δSH(ρ, ρ(r)), and we write f ∈ Mer(ρ, ρ(r)) and say “f is a meromorphic function of order ρ and normal type with respect to the proximate order ρ(r)”. For f ∈ Mer(ρ, ρ(r)) we use the following characteristics: T (r, f ), ρT [f ], σT [f, ρ(r)]. The charge of log |f | consists of integer positive and negative masses.

5.2 D -topology and topology of exceptional sets 5.2.1 Let α−mes be the Carleson measure defined in Section 2.5.4. Set for C ⊂ R2 , α − mesC := lim sup[α − mes(C ∩ KR )]R−α . R→∞

It is called the relative Carleson α-measure.

(5.2.1.1)

5.2. D -topology and topology of exceptional sets

153

Theorem 5.2.1.1 (Properties of the Relative Carleson Measure) One has rCm1) If C is bounded α − mesC = 0; α − mes(C1 ∩ C2 ) ≤ α − mesC1 + α − mesC2 ,

rCm2)

i.e., the relative Carleson measure is sub-additive; C1 ⊂ C2 ⇒ α − mes(C1 ) ≤ α − mesC2 ,

rCm3)

i.e., the relative Carleson measure is monotonic with respect to sets; α1 > α2 ⇒ α1 − mesC ≤ α2 − mesC,

rCm4)

i.e., the relative Carleson measure is monotonic with respect to α. Exercise 5.2.1.1 Prove this. A set C ⊂ R2 for which α − mesC = 0 is called a C0α − set. If α − mesC = 0 for all α > 0, C is called a C00 − set. Let us recall that if u1 ,u2 ∈ SH(ρ,ρ(r),R2 ), then u = u1 −u2 ∈ δSH(ρ,ρ(r),R2 ) (see Section 2.8.2). Theorem 5.2.1.2 (D -topology and Exceptional sets) Let u ∈ δSH(ρ, ρ(r), R2 ). In order that (5.2.1.2) ut → 0 in D as t → ∞ it is sufficient that u(z)|z|−ρ(|z|) → 0

(5.2.1.3)

as z → ∞ outside some C02 -set. If (5.2.1.2) holds, then (5.2.1.3) holds outside some C00 -set. 5.2.2 To prove Theorem 5.2.1.2 we need some auxiliary assertions. Recall that dz is an element of area following the notation of the previous chapters. 2 := C02 ∩ KR . Then Proposition 5.2.2.1 Let u ∈ SH(ρ, ρ(r), R2 ), and C0,R

 |u|(z)dz = o(Rρ(R)+2 )

(5.2.2.1)

2 C0,R

as R → ∞. Proof. Suppose (5.2.2.1) does not hold. Then there exists a sequence Rj → ∞ such that  −ρ(Rj )−2 |u|(z)dz = A > 0. (5.2.2.2) lim Rj Rj →∞

2 C0,R

j

154

Chapter 5. Applications to Entire Functions

Consider the following family of δ-subharmonic functions: −ρ(Rj )

uj (ζ) := Rj

u(ζRj ).

(5.2.2.2a)

It can be represented as a difference uj = u1,j − u2,j of subharmonic functions of the same form. Thus it is precompact in Lloc (Theorem 2.7.1.3). Let us choose a convergent subsequence for which we keep the same notation. Its limit v is a locally summable function. Now let χj be the characteristic functions of the sets 2 . Ej := Rj−1 C0,R j

Since mes Ej → 0 it is possible to choose a sequence (for which we keep the same notation) such that χj → 0 almost everywhere. We will also suppose that Rj are the same for χj and uj . Thus   |χj (ζ)uj (ζ) − 0 · v(ζ)|dζ = |χj (ζ)uj (ζ)|dζ → 0. |ζ|≤1

|ζ|≤1

By change of variables z = Rj ζ we obtain that   −ρ(Rj )−2 Rj |u|(z)dz = |χj (ζ)uj (ζ)|dζ → 0. 2 C0,R

|ζ|≤1

j

Hence the limit in (5.2.2.2) is equal to zero. Contradiction.



Proposition 5.2.2.2 Under condition (5.2.1.2) the set C := {z : |u(z)||z|−ρ(|z|) > } is a C00 -set for arbitrary . Proof. Assume the contrary; that is , ∃α > 0 such that α − mesC = 2δ > 0.

(5.2.2.3)

One can see that for some η > 0, lim sup(α − mes KηR )R−α ≤ δ/2. R→∞

Exercise 5.2.2.1 Check this.

(5.2.2.4)

5.2. D -topology and topology of exceptional sets

155

(5.2.2.3) and (5.2.2.4) imply that there exists a sequence Rj → ∞ such that lim α − mes[C ∩ (KRj \ KηRj )]Rj−α ≥

Rj →∞

3 δ. 2

Set Ej := Rj−1 C ∩ (KRj \ KηRj ). It is clear that Ej ⊂ K1 \ Kη and for sufficiently large j, α − mes Ej ≥ δ.

(5.2.2.5)

Set uj as in (5.2.2.2a). We claim that for large j and ζ ∈ Ej , |uj |(ζ) ≥

ρ |ζ| . 2

(5.2.2.6)

Indeed, |uj |(ζ) =

|u|(Rj ζ) ρ(R ) Rj j

=

|u|(z)

(1 + o(1))|ζ|ρ ≥ |ζ|ρ . ρ(|z|) 2 |z|

We used here properties of the proximate order and the equivalence z = Rj ζ ∈ C ∩ (KRj \ KηRj ) ⇔ ζ ∈ Ej . Exercise 5.2.2.2 Check this in detail. Now we will show that the condition (5.2.1.2) contradicts (5.2.2.6). Since u ∈ δSH(ρ, ρ(r), R2 ) it is a difference of u1 , u2 ∈ SH(ρ, ρ(r), R2 ). The corresponding sequences u1,j and u2,j are precompact in D and there exist subsequences (with the same notation) that converge to v1 and v2 , respectively. By Theorem 2.7.5.1 these sequences converge to v1 and v2 with respect to α − mes on K1 \ Kη . Since ut → 0 in D , it follows that v1 = v2 . Thus uj → 0 with respect to α−mes on K1 \Kη . However, this contradicts (5.2.2.5) and (5.2.2.6).  0 Proposition 5.2.2.3 Let {Cj }∞ 1 be a sequence of C0 -sets. There exists a sequence Rj → ∞ such that the set

C=

∞  j=1

is a C00 -set.

{Cj ∩ (KRj+1 \ KRj )}

(5.2.2.7)

156

Chapter 5. Applications to Entire Functions

Proof. Choose j ↓ 0 and αj ↓ 0. Set R0 := 1. Suppose Rj−1 was already chosen. Take Rj such that (5.2.2.8) αj − mes[C ∩ KRj−1 ] < j Rαj for R > Rj . It is possible because of property rC1) Theorem 5.2.1.1. We can also increase Rj so that (5.2.2.9) αj − mes[Cj ∩ KR ] < j Rαj and αj − mes[Cj+1 ∩ KR ] < j Rαj

(5.2.2.10)

for R > Rj . It is possible because Cj and Cj+1 are C00 -sets. Let us estimate αj −mes[C ∩KR ] for Rj ≤ R < Rj+1 . From (5.2.2.8), (5.2.2.9) and (5.2.2.10) we obtain αj − mes[C ∩ KR ] ≤ 3 j Rαj .

(5.2.2.11)

Let α > 0 be arbitrarily small. Find αj < α. For Rj+1 ≥ R > Rj we have α − mes[C ∩ KR ]R−α ≤ αj − mes[C ∩ KR ]R−αj ≤ 3 j . Hence α − mesC = 0.



5.2.3 Proof of Theorem 5.2.1.2. Let φ ∈ D(C) and supp φ ⊂ KR . Then for any > 0, ⎞ ⎛    ⎟ ⎜ J(t) := φ(z)ut (z)dz = ⎝ + ⎠ φ(z)ut (z)dz := J1 (t) + J2 (t). (5.2.3.1) KR \K

K

We have for J2 (see 2.8.2.3):  |J2 |(t) ≤ max |φ(z)| × const

T (r, |ut |)rdr ≤ const T ( , |ut |) 2 .

|z|≤

(5.2.3.2)

0

Further (see Theorem 2.8.2.1) T (r, |ut |) ≤ 2T (r, ut) + O(t−ρ(t) ) ≤ 2[T (r, u1,t ) + T (r, u2,t )] + O(t−ρ(t) ) ≤ 2[M (r, u1,t ) + M (r, u2,t )] + O(t−ρ(t) ).

(5.2.3.3)

Using (5.2.3.2), (5.2.3.3) and (3.1.2.3) we obtain lim sup |J2 (t)| ≤ const ρ+2 . t→∞

(5.2.3.4)

5.3. Asymptotic approximation of subharmonic functions

157

Exercise 5.2.3.1 Check this using the change of variable z = tζ. To estimate J1 (t) write    |u(z)|dz + |J1 (t)| ≤ const

|u(z)|dz t−ρ(t)−2 := J1,1 (t) + J1,2 (t),

2 C0,Rt

˜ t \C 2 K 0,Rt

(5.2.3.5) ˜ t := {z : t ≤ |z| ≤ Rt}. where K The summand J1,1 is o(1) as t → ∞ by (5.2.1.3). Exercise 5.2.3.2 Check this using the properties of the proximate order (Theorem 2.8.1.3, ppo3). The summand J1,2 is o(1) by Theorem 5.2.2.1. Thus lim sup |J(t)| ≤ const ρ+2 t→∞

for any . Hence it is equal to zero and the sufficiency of (5.2.1.3) has been proved. Let us prove sufficiency of (5.2.1.2). Let j ↓ 0. By Theorem 5.2.2.2 we choose a C00 -set Cj outside which |u(z)||z|−ρ(|z|) < j . We construct the set C by (5.2.2.7). Outside C we have (5.2.1.3). And by  Theorem 5.2.2.3 it is a C00 -set.

5.3 Asymptotic approximation of subharmonic functions 5.3.1 One of the widely applied methods of constructing entire functions with a prescribed asymptotic behavior is the following: First construct a subharmonic function behaving asymptotically as the logarithm of modulus of the entire function,and then approximate it in some sense by the logarithm of modulus of entire function such that the asymptotic is preserved. Various queries about the a precision of the approximation and about the metric in which it was implemented generated a spectrum of theorems of such kind that we will demonstrate. Historically the first theorems of this kind were proved for concrete functions, the masses of which were concentrated on sufficiently smooth curves (in particular, on lines, see, e.g., [BM, Ev, Kj, Ar], . . . ) In such cases the approximation was very precise and exceptional sets where the approximation failed were small and determined. The first general case was proved in [Az(1969)]. Next it was developed in [Yu(1982)], and vastly improved in [Yu(1985)]. It is the following

158

Chapter 5. Applications to Entire Functions

Theorem 5.3.1.1 (Yulmukhametov) Let u ∈ SH(ρ). Then there exists an entire function f such that for every α ≥ ρ, |u(z) − log |f (z)|| < Cα log |z| for z ∈ / Eα , where Eα is an exceptional set that can be covered by discs Dzj ,rj := {z : |z − zj | < rj } satisfying the condition 

rj = o(Rρ−α ), R → ∞.

|zj |>R

This theorem is precise in the following sense: If ||z| − log |f (z)|| = o(log |z|), z → ∞, z ∈ / E, then for every covering of E by discs Dzj ,rj and every > 0 

rj ≥ R1− , R → ∞,

|zj |
i.e., in any case this sum is not even bounded. However it is necessary to remark that the construction from [Yu(1985)] “ rigidly ” fastens zeros of the entire function, whereas the construction of [Az(1969)] and [Yu(1982)] gives some possibilities to move them, which is needed in some constructions. Let us also mention that such approximation generates an approximation of a plurisubharmonic function by the logarithm of the modulus of an entire function in Cp (see [Yu(1996)]). It is also useful to approximate subharmonic functions in an integral metric, for example Lp , as was done in [GG]. Set  2π 1/p p

g p := |g(t)| dt . 0

Denote by Q(r, u) a function that satisfies the conditions: 1) if u is of finite order, then Q(r, u) = O(log r); 2) if u is of infinite order, then Q(r, u) = O(log r + log μu (r)). Theorem 5.3.1.2 (Girnyk, Gol dberg) For every subharmonic function in C u there exists an entire function f such that u(rei· ) − log |f |(rei· ) p = Q(r, u). This theorem also considers functions of infinite order. In this case, it is possible to replace μu (r) by T (r, u) or M (r, u) in Q(r, u) outside an exceptional set E ⊂ R+ of finite measure. This theorem is also unimprovable for subharmonic

5.3. Asymptotic approximation of subharmonic functions

159

functions of finite order, because, for example, u = 12 log |z| gives, as it is possible to prove:

u(rei• ) − log |f |(rei• ) p > 0. lim inf r→∞ log r However it was found [LS], [LM] that the remainder term O(log |z|) that was regarded the best possible is not precise and in some “regular” cases can be replaced with O(1) outside a bigger (but still “small”) set . Set, for E ⊂ C: Δ(E) := lim sup r→∞

mes E ∩ D0,R . R2

Theorem 5.3.1.3 (Lyubarskii, Malinnikova) Let u be a subharmonic function in C with μu satisfying the conditions: μu (C) = ∞ and there exists α > 0, q > 1, R0 > 0 such that μu (D0,qR \ D0,R ) > α for all R > R0 . Then there exists an entire function f such that for every > 0, |u(z) − log |f (z)|| < C for z ∈ C \ E with Δ(E ) < . So if μu has no “Hadamard’s gaps” such approximation is possible. In this book we restrict ourself to a weaker and simply proved theorem that is sufficient for our aim Theorem 5.3.1.4 (Approximation Theorem) For every u ∈ SH(ρ, ρ(r)) there exists an entire function f such that D − lim (u − log |f |)t = 0. t→∞

Nevertheless this theorem has an important Corollary 5.3.1.5 For every u ∈ SH(ρ, ρ(r)) there exists an entire function f such that Fr[u] = Fr[f ]. 5.3.2 Now we prove Theorem 5.3.1.4. We can suppose, because of Theorem 3.1.6.1 (Dependence Fr on V• ), that in the definition of (•)t (see 3.1.2.1) Vt ≡ I We prove this theorem for the case non-integer ρ. For proving this theorem we need Lemma 5.3.2.1 Let u ∈ δSH(ρ, ρ(r)), f or ρ non-integer, and ν is its charge. Then ut → 0 iff νt → 0 in D as t → ∞.

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Chapter 5. Applications to Entire Functions

Proof. Sufficiency. Suppose ut := (u1 )t − (u2 )t → 0. There exists a subsequence tj → ∞ and subharmonic functions v1 and v2 such that utj = (u1 )tj − (u2 )tj → v1 − v2 := v = 0.

(5.3.2.1)

Applying to (5.3.2.1) the continuity of Δ in D and using the conditions of the theorem, we obtain 1 νtj → Δv = 0. 2π Hence v is harmonic. Since v1 , v2 ∈ U [ρ, ] also v ∈ U [ρ] (see Theorem 2.8.2.1, t3), t4) and Theorem 2.8.2.3). Exercise 5.3.2.1 Prove this in detail. By Theorem 3.1.4.3 we obtain v = 0. Contradiction. Necessity. Since the Laplace operator is continuous in D -topology, the asser1 tion ut → 0 implies νt := 2π Δut → 0.  Now we describe a construction of the zero distribution of the future entire function. Let u ∈ SH(ρ) and μ be its mass distribution. Set Rj+1 := Rj (j + 1)4/κ

(5.3.2.2)

where κ := min(ρ − [ρ], [ρ] + 1 − ρ). Let us divide all the plane by circles of the form SRj := {|z| = Rj } such that Rj+1 /Rj → ∞ and μ(SRj ) = 0. Exercise 5.3.2.2 Prove that it is possible. Choose a sequence δj ↓ 0. Divide every annulus Kj := {z : Rj ≤ |z| < Rj+1 } by circles SRj,n for  Rj,n := where

1 + δj 1 − δj

n Rj , n = 0, 1, 2, . . . , nj , ⎡

nj := ⎣

log log

Rj+1 Rj 1+δj 1−δj

⎤ ⎦,

and by rays Lk := z : arg z = kδj , k = 0, 1, . . . , [2π/δj ]. They divide all the plane into sectors Kj,n,k . We can choose δj in such a way that μ(∂Kj,n,k ) = 0 because μ(Kj,n,k ) is a monotonic function of δj and has only a countable set of jumps. Exercise 5.3.2.3 Explain this in detail.

5.3. Asymptotic approximation of subharmonic functions

161

Choose a point zj,n,k in every sector Kj,n,k and concentrate all the mass of the sector at this point. In other words we consider a new mass distribution μ ˆ that ˆ(zj,n,k ) = μ(Kj,n,k ). has masses concentrated in the points zj,n,k and μ The next lemma shows that μ ˆ is close to μ. Lemma 5.3.2.2 One has μ ˆt − μt → 0 in D as t → ∞. Proof. Assume the contrary, i.e., μ ˆ t − μt → 0. Choose a sequence tl → ∞ such that μ ˆtl → νˆ and μtl → ν, ν, νˆ ∈ M[ρ], ν = νˆ. Then there exists a disc Kz0 ,r0 := {z : |z − z0 | < r0 } such that ν(Kz0 ,r0 ) = νˆ(Kz0 ,r0 ). We can assume that this disc does not contain zero since for all the ν ∈ M[ρ] the condition ν(Kr ) ≤ Δrρ , ∀r > 0 is fulfilled. Suppose, for example, ν(Kz0 ,r0 ) > νˆ(Kz0 ,r0 ).

(5.3.2.3)

Set a := ν(Kz0 ,r0 ) − νˆ(Kz0 ,r0 ) > 0. Choose such that ν(Kz0 ,r0 ) < ν(Kz0 ,r0 − ) + a/3.

(5.3.2.4)

This is possible because the countable additivity of νˆ implies lim ν(Kz0 ,r ) =  r ↑r

ν(Kz0 ,r ). Consider now the sets tl Kz0 ,r0 , tl Kz0 ,r0 − . For sufficiently large tl they are contained in the union of the annuluses Kjl ∪ Kjl +1 . As jl → ∞ the diameters of all the sectors Kjl ,n,k are o(Rjl ) uniformly. Thus they are o(tl ). Hence for such tl ’s we can find a union Γl of sectors covering tl Kz0 ,r0 − that does not intersect the circle of tl Kz0 ,r0 . We have μ ˆ(Γl ) = μ(Γl ) by definition of μ ˆ. Using the monotonicity of measures, we obtain μ(tl Kz0 ,r0 − ) ≤ μ ˆ (tl Kz0 ,r0 ), whence μtl (Kz0 ,r0 − ) ≤ μ ˆtl (Kz0 ,r0 ). Passing to the limit as l → ∞ and using Theorems 2.2.3.1 and 2.3.4.4, we obtain ν(Kz0 ,r0 − ) ≤ νˆ(Kz0 ,r0 ). Using (5.3.2.4), we obtain ν(Kz0 ,r0 ) − 1/3[ν(Kz0 ,r0 ) − νˆ(Kz0 ,r0 )] ≤ νˆ(Kz0 ,r0 ) and hence ν(Kz0 ,r0 ) ≤ νˆ(Kz0 ,r0 ), that contradicts (5.3.2.3). Since ν and νˆ are symmetric in this reasoning the lemma is proved.  Let us finish the proof of Theorem 5.3.1.4 for non-integer ρ. We construct a distribution n with integer masses concentrated at points zj,k,n . Set n(zj,k,n ) := [ˆ μ(zj,k,n )]

162

Chapter 5. Applications to Entire Functions

and estimate the growth of the difference δμ := μ ˆ−n that is also a mass distribution concentrated at the same points. Since δμ(zj,k,n ) ≤ 1 it is sufficient to count the number of points in the disc KR . The number of points in the annulus {Rj ≤ |z| < R} is found from (5.3.2.2),



−1 1 + δj 2π R δμ({Rj ≤ |z| < R}) ≤ log log 1 − δj δj Rj log(j + 1) ≤ const × = const ×(j + 1)4 log(j + 1). δj2 The mass of the disc KR is estimated by the inequality δμ(KR ) ≤ const ×

n−1 

(k + 1)4 log(k + 1) = o(n6 ) = o(R )

(5.3.2.5)

k=0

for any > 0 because R > Rn−1 = ((n − 1)!)4/κ . Exercise 5.3.2.4 Check this in detail. The estimate (5.3.2.5) shows that δμt → 0

(5.3.2.6)

as t → ∞. Lemma 5.3.2.2 and (5.3.2.6) imply that μt − nt → 0.

(5.3.2.7)

Set u1 (z) := Π(z, n, p) (see (2.9.2.1)) where Π is a canonical potential. This is a subharmonic function in the plane with integral masses. Thus it is the logarithm of the modulus of the entire function  f (z) = E(z/zj,k,n ). (5.3.2.7) implies by Lemma 5.3.2.1 that ut − (u1 )t → 0 and this is the assertion of Theorem 5.3.1.4 for non-integer ρ. 

5.4 Lower indicator of A.A. Gol dberg

163

5.4 Lower indicator of A.A. Gol dberg. Description of lower indicator Description of the pair: indicator-lower indicator 5.4.1 Now we consider the lower indicator. For an entire function of finite order ρ and normal type it can be defined in one of the following ways: h1 (φ, f ) := sup {

lim inf

/ C∈C reiφ →∞,reiφ ∈C

log |f (reiφ )|r−ρ(r) },

(5.4.1.1)

where C is the set of C0 -sets (see [Le, Ch. II, § 1]), i.e., the sets that can be covered by a union of discs Kδj (zj ) := {z : |z − zj | < δj } such that 1  lim δj = 0. R→∞ R |zj |
The exclusion of C0 -sets is necessary because we must exclude from our consideration some neighborhoods of roots of f (z) where log |f (z)| is near −∞. Similarly, define h2 (φ, f ) := sup {

lim inf

/ E(φ)∈E r→∞,r ∈E(φ)

log |f (reiφ )|r−ρ(r) },

(5.4.1.2)

where E is the set of E0 -sets (see [Le, Ch. III]), i.e., sets E ⊂ (0, ∞] satisfying the condition lim mes{E ∩ (0, R)}R−1 = 0. R→∞

The definition (5.4.1.1) was introduced by A.A. Gol dberg (see [Go(1967)]). We will use the definition (3.2.1.2) h(φ, f ) = inf{v(eiφ ) : v ∈ Fr[f ]}.

(5.4.1.3)

It was proved in [AP, Thm. 1] that the definitions (5.4.1.1), (5.4.1.2) and (5.4.1.3) coincide. Let us note that (5.4.1.3) uses the definition (3.2.1.2) only on the circle {|z| = 1}. However, it is easy to check, by using Theorem 3.2.1.2 that for h(z) = |z|ρ h(arg z) properties h1) and h2), Theorem 3.2.1.1, are preserved. Exercise 5.4.1.1 Check this. We are going to prove Theorem 5.4.1.1 Let g(φ) be a 2π-periodic function that is either semicontinuous from above or ≡ −∞ and ρ(r) → ρ be an arbitrary approximate order. Then there exists an entire function f ∈ A(ρ, ρ(r)) such that h(φ, f ) = g(φ) for all φ ∈ [0, 2π).

(5.4.1.4)

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Chapter 5. Applications to Entire Functions

5.4.2 We will use the following assertion that is a corollary of Theorem 4.3.5.1 and Corollary 5.3.1.5: Theorem 5.4.2.1 Let Λ ⊂ U [ρ] be a compact, connected and T• -invariant subset. Then for any proximate order ρ(r) → ρ there exists f ∈ A(ρ, ρ(r)) such that h(φ, f ) = sup{v(eiφ ) : v ∈ Λ},

(5.4.2.1)

h(φ, f ) = inf{v(e ) : v ∈ Λ}.

(5.4.2.2)

iφ

Exercise 5.4.2.1 Prove Theorem 5.4.2.1. For the sake of clarity let us restrict ourselves to non-integer ρ. We will construct a set Λ such that inf{v(eiφ ) : v ∈ Λ} = g(φ). Denote H(z, p) := log |1 − z| + 

p  zk k=1

k

; p = [ρ],

γ(z, K, λ) := −λ + K|z − 1|, λ, K ≥ 0. Note the following properties of these functions: a)

min δH(z, p)|z|−ρ → 0, as δ → 0;

|z−1|≥δ

b) δH(z, p)|z|−ρ ≤ Aδ, for all z ∈ C, where A depends only on p; c)

max

|z−1|≤0.5K

1 γ(z, K, λ) ≤ − . 2

(5.4.2.3)

Exercise 5.4.2.2 Prove properties a), b), c). Let us note that H(1, p) = −∞. Consider the family: Λ∞ = {vθ,τ (z) := H(ze−iθ τ, p)τ −ρ : θ ∈ [0, 2π), τ ∈ (0, ∞)} ∪ 0. This family is contained in U [ρ] because of b) and closed in D -topology. It is also T• -invariant, hence, satisfies the conditions of Theorem 5.4.2.1. For every φ ∈ [0, 2π) there exists θ0 (= φ), and τ0 (= 1) such that vθ0 ,τ0 (eiφ ) = H(1, p) = −∞ Hence inf{v(eiφ ) : v ∈ Λ∞ } = −∞. (5.4.2.4) For the general case this construction will be improved, cutting the “trunk” of the function H(ze−iθ , p).

5.4. Lower indicator of A.A. Gol dberg

165

Take δ small enough so that the following conditions hold: 1 δH(z, p)|z|−ρ ≥ − , for |z − 1| ≥ δ, 4 1 δH(z, p) ≥ − , for |z − 1| = δ, 4 1 . δ≤ 2K

(5.4.2.5) (5.4.2.6) (5.4.2.7)

Then δH(z, p) > γ(z, K, λ), for|z − 1| = δ. Denote



W (z, K, δ, λ) :=

max{δH(z, p), γ(z, K, λ)}, for |z − 1| < δ, δH(z, p), for |z − 1| ≥ δ.

(5.4.2.8)

(5.4.2.9)

Lemma 5.4.2.2 The following holds: aw) The function W (z, K, δ, λ) is subharmonic in C; bw) supp μW  {|z − 1| < δ}; cw) sup W (z, •, δ, λ)|z|−ρ ≤ Aδ,

(5.4.2.10)

z∈C

where A depends only on p. Proof. For |z − 1| < δ, W is subharmonic as the maximum of two subharmonic functions. For |z − 1| ≥ δ it is harmonic even in the neighborhood of the circle |z − 1| = δ, because of inequality (5.4.2.8). So aw) and bw) hold. The assertion cw) follows from b) and c) (5.4.2.3) above. Now we get to the proof of (5.4.1.4). Let gn ↓ g be a sequence of continuously differentiable functions that converges to g monotonically. This is possible, because g is semicontinuous from above. Exercise 5.4.2.3 Prove that Theorem 2.1.2.9 and the Weierstrass theorem of approximation of every periodic function by trigonometrical polynomials imply the last assertion. We write Mn := max gn+ (φ) φ

+

where as usual a = max(a, 0). Set vθ,n (z) := W (ze−iθ , Kn , δn , Mn + 1 − gn (θ)) + (Mn + 1)|z|ρ , where δn is chosen small and Kn is chosen large. Set z = τ eiφ . It is clear that vφ,n (eiφ ) = gn (φ) for all Kn , δn .

(5.4.2.11)

166

Chapter 5. Applications to Entire Functions

We can choose Kn so large and δn so small that γ(z, Kn, Mn + 1 − g(θ))|z|−ρ ≥ gn (φ) for |z − 1| ≤ δn , because gn has bounded derivative. After that we can make δn smaller so that for |z − 1| ≥ δn the inequality (5.4.2.8) would hold. Exercise 5.4.2.4 Estimate exactly Kn and δn via the derivative of gn . Then

vθ,n (z)|z|−ρ ≥ gn (φ)

for all z = reiφ . Thus

min vθ,n (τ eiφ )τ −ρ = gn (eiφ ), θ,τ

and the minimum is attained for τ = 1, θ = φ. Let us note that from (5.4.2.10) we have sup sup vθ,n (z)|z|−ρ ≤ Aδn + Mn + 1 ≤ A + M1 + 1. θ

z∈C

Consider now the family of functions Λ0 := {vθ,n (zτ )|τ |−ρ : θ ∈ [0; 2π), n = 1, 2, . . . , τ ∈ (0; ∞)}. It is contained in U [ρ, σ] for σ = A + M1 + 1 and is T• -invariant. Let Λ be its closure in D . Let us show that g(φ) = inf{v(eiφ ) : v ∈ Λ}.

(5.4.2.12)

Indeed, for every sequence vj ∈ Λ1 vj (eiφ ) ≥ inf gn (φ) = g(φ). n

Let v ∈ Λ. By Theorem 2.7.4.1 (D and Quasi-everywhere Convergence) v(z) := (D − lim vj )(z) = (lim sup vj )∗ (z). j→∞

j→∞

Hence v(eiφ ) ≥ g(φ). However, the infimum is attained for every φ on the sequence vφ,n (z) because of (5.4.2.11). Hence (5.4.2.12) holds and Theorem 5.4.1.4 is proved.  5.4.3 Now we describe the pair: indicator-lower indicator. Let h be a 2π-periodic, ρ-trigonometrically convex function (ρ-t.c.f) and let g be a 2π-periodic upper semicontinuous function. Further they are indicator and lower indicator of an entire function, and hence must satisfy the condition h(φ) ≥ g(φ), φ ∈ [0, 2π).

(5.4.3.1)

5.4. Lower indicator of A.A. Gol dberg

167

An interval (a, b) ⊂ [0, 2π) is called a maximal interval of ρ-trigonometricity of the function h if h(φ) = A cos ρφ + B sin ρφ, φ ∈ (a, b) (5.4.3.2) for some constants A, B, and h has no such representation on any larger interval (a , b ) ⊃ (a, b). A function h is said to be strictly ρ-t.c.f. if it is a ρ-t.c.f. and is not ρtrigonometrical on any interval. If the function h is a strictly ρ-t.c.f., then h and g (satisfying other previous bounds) could be an indicator and lower indicator of an entire function f ∈ A(ρ(r)). However this is not so if the function h has an interval of trigonometricity. Recall, for example, the famous M. Cartwright Theorem [Le, Ch. IV, § 2, Thm. 6]: if an indicator of an entire function is trigonometrical on an interval (a, b) with b − a > π/ρ, then the function is a CRG -function on this interval, i.e., h(φ) = g(φ), φ ∈ (a, b).

(5.4.3.3)

Let us formulate all the necessary conditions of such kind. Let (a, b) be a maximal interval of ρ-trigonometricity of the function h. The M. Cartwright theorem can be formulated as the implication (b − a > π/ρ) ⇒ (5.4.3.3).

(5.4.3.4)

The following implications are also necessary: (∃φ0 ∈ (a, b) : h(φ0 ) = g(φ0 )) ⇒ (5.4.3.3),

(5.4.3.5)

(h(a) = g(a) ∧ h+ (a) = h− (a)) ⇒ (5.4.3.3), (h(b) = g(b) ∧ h+ (b) = h− (b)) ⇒ (5.4.3.3),

(5.4.3.6a) (5.4.3.6b)

where h± (a) and h± (b) are the right and left derivatives of the function h at the points a and b. (b − a = π/ρ ∧ h+ (a) = h− (a)) ⇒ (5.4.3.3),

(5.4.3.7a)

(b − a = π/ρ ∧ = ⇒ (5.4.3.3),  h(φ) − g(φ) = 0 ⇒ (5.4.3.3), lim inf φ→a+0 φ−a  h(φ) − g(φ) = 0 ⇒ (5.4.3.3). lim inf φ→b−0 b−φ

(5.4.3.7b)

h+ (b)

h− (b)

(5.4.3.8a) (5.4.3.8b)

Now we shall give an exact formulation. The functions h and g are said to be concordant if at least one of the following conditions holds: 1. h is strictly ρ-t.c.; 2. for each (a, b) that is a maximal interval of ρ-trigonometricity of the function h the implications (5.4.3.4)–(5.4.3.8b) are satisfied.

168

Chapter 5. Applications to Entire Functions

Theorem 5.4.3.1 Let 0 < ρ < ∞, h(φ) be a 2π-periodic, ρ-t.c.f., g(φ) be an upper semicontinuous, 2π-periodic function, h(φ) ≥ g(φ) for all φ, and h ≡ g. A function f ∈ A(ρ(r)) which simultaneously satisfies the identity hf ≡ h, hf ≡ g with an arbitrary proximate order ρ(r) → ρ exists if and only if the functions h and g are concordant. 5.4.4 Proof of necessity. Note that implication (5.4.3.4) is a corollary of (5.4.3.6a) or (5.4.3.6b), because every ρ-trigonometrical function is continuous and has continuous derivative in (a, b). Recall that (•)[t] was defined by (3.1.2.4a). From properties of the limit set Fr[f ] (Theorem 3.1.2.2, fr2), fr3)) and the definition of indicators ((3.1.2.1), (3.1.2.2)) we can obtain for every function v ∈ Fr[f ] the inequality v(τ eiφ ) ≤ τ ρ h(φ), φ ∈ [0, 2π), τ > 0.

(5.4.4.1)

Since h(φ) is ρ-trigonometrical for φ ∈ (a, b), the function H(reiφ ) := rρ h(φ) is harmonic in the angle Γ(a, b) := {reiφ : φ ∈ (a, b), r ∈ (0, ∞)}, whence the function v − H is subharmonic and nonpositive in Γ(a, b). By virtue of the maximum principle, either v < H in Γ(a, b) or v ≡ H in Γ(a, b) for each v ∈ Fr[f ]. Note that the condition v ≡ H in Γ(a, b) implies v ≡ H in Γ[a, b] for the closed interval because of the upper semicontinuity of v. Let us prove (5.4.3.5). For every v ∈ Fr[f ] we have v(reiφ0 ) − H(reiφ0 ) = 0 whence by the maximum principle v = H in Γ(a, b). Hence (5.4.3.3) holds. Let us prove (5.4.3.6a). Assume the contrary:h(a) = g(a) ∧ h+ (a) = h− (a) holds, but there exists φ0 ∈ (a, b) such that h(φ0 ) > g(φ0 ). Then there exists v ∈ Fr[f ] such that g(φ0 ) ≤ v(eiφ0 ) < h(φ0 ) whence v(τ eiφ ) < τ ρ h(φ) ∈ Γ(a, b).

(5.4.4.2)

Without loss of generality, we can assume that v(z) > −∞, otherwise we can replace v with max(v, −C) for a large positive constant C > 0. We choose 0 < τ1 < τ2 and to every function Wj (reiφ ) := v[τj ] (reiφ+a ) − rρ h(φ + a), j = 1, 2, γ = b − a, reiφ ∈ Γ(0, γ) we apply the following lemma due to A.E. Eremenko and M.L. Sodin [So] (see also [PW, Ho]):

5.4. Lower indicator of A.A. Gol dberg

169

Lemma 5.4.4.1 (E.S.) Let W be a subharmonic nonpositive function inside the angle Γ(0, γ), γ > 0. Then the following implication is valid, % $ W (eiφ ) = 0 ⇒ W ≡ 0. lim sup φ φ→0 If the condition of this theorem is not satisfied for W ∗ (reiφ ) = max v[τ ] (reiφ ) τ ∈[τ1 ,τ2 ]

it would be possible to insert a ρ-t.c.function between h(φ) − (φ − a) (for a small

) and v(eiφ ). However, such a function does not exist, because of the negative jump of the derivative. So it will be a contradiction. See further for details. From Lemma 5.4.4.1 we get lim inf

h(φ) − v[τ1 ] (eiφ ) := α1 > 0 φ−a

lim inf

h(φ) − v[τ2 ] (eiφ ) := α2 > 0. φ−a

φ→a+0

and likewise φ→a+0

So a Δ > 0 can be chosen such that a + Δ < b and the inequalities H(τj eiφ ) − v[τj ] (eiφ ) > ατjρ (φ − a), j = 1, 2,

(5.4.4.3)

where α := 1/2 min(α1 , α2 ), hold for all φ ∈ [a, a + Δ]. We write β := min (H(τ ei(a+Δ) ) − v(τ ei(a+Δ) )) τ ∈[τ1,τ2 ]

which is positive because of (5.4.4.2). Let us choose > 0 small enough to

< min(α, β(τ2 )−ρ Δ−1 )

(5.4.4.4)

and let us consider the ρ-trigonometrical function h (φ) := ρ−1 (h (a) − ) sin ρ(φ − a) + h(a) cos ρ(φ − a), φ ∈ (a, b) that coincides with h(φ) = ρ−1 h (a) sin ρ(φ − a) + h(a) cos ρ(φ − a), φ ∈ (a, b) in the point φ = a but has a tangent that is lower than the tangent of h. Further h(φ) − h (φ) = ρ−1 sin ρ(φ − a) ≤ (φ − a), φ ∈ [a, a + Δ].

(5.4.4.5)

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Chapter 5. Applications to Entire Functions

Combining (5.4.4.3)–(5.4.4.5) we obtain v[τj ] (eiφ ) < τj ρ h(φ) − α(φ − a) < τj ρ h(φ) − (φ − a)

(5.4.4.6)

≤ τj h (φ), φ ∈ [a, a + Δ], j = 1, 2, ρ

v(τ e ) ≤ τ ρ h (a + Δ) + τ ρ Δ − β iφ

(5.4.4.7)

< τ h (a + Δ), τ ∈ [τ1 , τ2 ]. ρ

We write G := {reiφ : φ ∈ [a, a + Δ], τ ∈ [τ1 , τ2 ]}.

(5.4.4.8)

It follows from (5.4.4.6), (5.4.4.7) that v(reiφ ) < rρ h (φ), reiφ ∈ ∂G, where ∂G is the boundary of the domain G. Since the functions v(reiφ ) and rρ h (φ) are subharmonic in G, by virtue of the maximum principle we have v(reiφ ) < rρ h (φ), reiφ ∈ G.

(5.4.4.9)

Let us consider the function H1 (reiφ ) := rρ h1 (φ), reiφ ∈ Γ(a − Δ, a + Δ) 

where h1 (φ) :=

h(φ), φ ∈ (a − Δ, a], h (φ), φ ∈ [a, a + Δ).

The function H1 is continuous in Γ(a − Δ, a + Δ) and subharmonic in the angles Γ(a − Δ, a) and Γ(a, a + Δ). Let us prove that it is subharmonic at the point z = eia . Let M(z, R, v) be the mean value of v over the circle {ζ : |ζ − z| = R} (see (2.6.1.1)). Taking into consideration (5.4.4.9) and subharmonicity of v (see (2.6.1.1)), for all small R we have M(eia , R, H1 ) ≥ M (eia , R, v) ≥ v(eia ) = H1 (eia ). Hence H1 is subharmonic for z = eia . Since H1 is homogeneous, i.e., H1 (kz) = k ρ H1 (z), M(keia , kR, H1 ) = k ρ M(eia , R, H1 ) ≥ k ρ H1 (eia ) = H1 (keia ). So H1 is subharmonic on the ray {z = keia : k ∈ (0, ∞)} and hence in the angle Γ(a − Δ, a + Δ). Thus h1 (φ) is a ρ-t.c.f. for φ ∈ (a − Δ, a + Δ). However, by construction (h1 )− (a) = h− (a) = h+ (a) = (h )+ + = (h1 ) (a) + and this contradicts the fact that h1 is ρ-t.c.f. Concordance of the implication (5.4.3.6a) is proved.

5.4. Lower indicator of A.A. Gol dberg

171

5.4.5 Here we continue the proof of necessity. Pass to the proof of necessity of the condition (5.4.3.7a). Assume the contrary. Then there exists v ∈ Fr[f ] and φ0 ∈ [a, b] such that g(φ0 ) ≤ v(eiφ0 ) < h(φ0 ), whence by virtue of the maximum principle, v(τ eiφ ) < τ ρ h(φ) for τ eiφ ∈ Γ(a, b). Actually v(τ eiφ ) ≤ τ ρ h(φ) everywhere and on the circle we have strict inequality. If v(τ eia ) = H(τ eia ) for a τ > 0, then v[τ ] (eia ) = h(a), and it will suffice to repeat the arguments used in proving (5.4.3.6a) with v[τ ] instead of v. Exercise 5.4.5.1 Do that. So it is sufficient to examine the case v(τ eia ) < H(τ eia ), τ > 0. Denote T (φ) := h (a)ρ−1 sin ρ(φ − a) + h(a) cos ρ(φ − a). This is a ρ-trigonometrical function, the graph of which is tangent to the graph of h(φ) at the point a. There are two possibilities for T (φ) on some small interval φ ∈ (a−γ, a), γ > 0 : either T (φ) < h(φ) or T (φ) = h(φ). Inequality T (φ) > h(φ) contradicts ρ-t.convexity at the point a. The equality on the sequence of points φj → a − 0 contradicts the maximum principle for ρt.c.functions. Exercise 5.4.5.2 Why is it? If T (φ) = h(φ), φ ∈ (a − γ, a), then h is ρ-trigonometrical on the interval (a−γ, b) ⊃ (a, b) that was already considered in the case (5.4.3.4) (M. Cartwright’s Theorem). So we assume T (φ) < h(φ), φ ∈ (a − γ, a). We set h1 (φ) := h(φ) − T (φ), φ ∈ (a − γ, a), v1 (reiφ ) := v(reiφ ) − rρ T (φ), reiφ ∈ Γ(a − γ, b). Then h1 (φ) = 0 for φ ∈ [a, b], h1 (φ) > 0 for φ ∈ (a − γ, a) and h (a) = 0. The function v1 (eiφ ) < 0, φ ∈ [a, b). Let us analyze the behavior of the function v1 (eiφ ) at the point b. Either v1 (eib ) < 0 or v1 (eib ) = 0 but lim sup v1 (eiφ )(b − φ)−1 ≤ −C φ→b−0

for some C > 0 by Lemma 5.4.4.1 (E.S.). From the other side v1 (eiφ ) is strictly negative also in some left (say, (a − Δ, a)) neighborhood of a because of upper semicontinuity. In any case v1 (eiφ ) can be majorated on the interval (a − Δ, b) by the function h := −A sin(ρ − )(b − φ) with sufficiently small A. A point of intersection of the graph of h with the axis 0, φ can be regulated by and can be chosen so close to the point a that the graph of h also intersects the graph of h1 (φ), at some point θ0 < a because h1 (a) = h1 (a) = 0.

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Exercise 5.4.5.3 Make the precise proof with all the estimates. Let the parameters A, , θ0 be fixed as above. Denote S := {reiφ : φ ∈ (θ0 , b), 0 < r < 1}. Then H (reiφ ) := rρ− h (φ) is harmonic in the sector S and satisfies the inequality H (reiφ ) ≥ v1 (reiφ ) on ∂S. Hence H (reiφ ) ≥ v1 (reiφ ) on S. Thus v(reiφ ) ≤ H(reiφ ) + H (reiφ ), reiφ ∈ S.

(5.4.5.1)

Let M(r, v) be the mean value of the function on the circle {ζ : |ζ| = r} (see 2.6.1.1). Using (5.4.5.1) we have 

b M(r, v) ≤

iφ

iφ

H(reiφ )dφ

[H(re ) + H (re )]dφ + θ0

[0,2π)\(θ0 ,b)

≤ d1 r − d2 r ρ

ρ−

,

d1 , d2 > 0.

So we get M(r, v) < 0 = v(0) for sufficiently small r > 0 which contradicts the subharmonicity of the function v at zero. 5.4.6 Now we complete proof of necessity, proving (5.4.3.8a,b). Assume the contrary: suppose h(φ) − g(φ) lim inf =0 (5.4.6.1) φ→a+0 φ−a but there exists a φ0 ∈ (a, b) such that h(φ0 ) > g(φ0 ). Then there exists a function v ∈ Fr[f ] such that g(φ0 ) ≤ v(eiφ0 ) < h(φ0 ). (5.4.6.2) Then the function v1 (reiφ ) := v(reiφ ) − H(reiφ ) is subharmonic and nonpositive in Γ(a, b). By virtue of the maximum principle v1 (reiφ ) < 0, reiφ ∈ Γ(a, b). From (5.4.6.1) we obtain 0 = lim inf

φ→a+0

h(φ) − g(φ) h(φ) − v(eiφ ) v1 (eiφ ) ≥ lim inf = − lim inf φ→a+0 φ→a+0 φ − a φ−a φ−a

whence, recollecting that v1 (eiφ ) < 0, we get lim sup φ→a+0

v1 (eiφ ) = 0. φ−a

Applying Lemma 5.4.4.1 (E.S.) to the function W (reiφ ) = v1 (reiφ+a ), reiφ ∈ Γ(0, γ), γ = b − a we get v1 ≡ 0 in Γ(a, b) which leads to a contradiction. The implication (5.4.3.8b) is proved in the same way. So the proof of necessity in Theorem 5.4.3.1 is completed.  We do not include here the proof of sufficiency and refer the readers to the original paper [Po(1992)].

5.5. Asymptotic extremal problems. Semiadditive integral

173

5.5 Asymptotic extremal problems. Semiadditive integral 5.5.1 Suppose some class of entire functions is determined by asymptotic behavior of their zeros, and we want to know what is the restriction on asymptotic behavior of functions: for example, to estimate the indicator of such a function. The first example of such a problem was considered by B.Ya. Levin in [Le, Ch. IV, § 1, Example]. A developed theory of such estimates was constructed in the papers of A.A. Gol dberg [Go(1962)] and his pupils [Kon], [KF]. We consider this theory from the point of view of limit sets. Let M  M(ρ) (see (3.1.3.4)) be a convex set of measures which is closed in D and is invariant with respect to the transformation (•)t (see (3.1.3.1), (3.1.3.2)) and let A(M) be a class of entire functions f for which Fr[nf ] ⊂ M. We suppose ρ is non-integer. Recall that canonical potential Π(z, ν, p) is defined by: (see (2.9.2.1))  Π(z, ν, p) := Gp (z/ζ)ν(dζ), C

where ν is a measure and Gp (z) := log |1 − z| + 

p  zk k=1

k

.

Theorem 5.5.1.1 [AP] The relation h(φ, f ) = sup{Π(eiφ , ν, p) : ν ∈ M}

(5.5.1.1)

is valid. There exists f ∈ A(M) for which the equality holds in (5.5.1.1) for all φ. Proof. We should only prove that there exists an entire function with such indicator. Consider the set Λ := {Π(eiφ , ν, p) : ν ∈ M}. It is a convex set contained in U [ρ]. Thus there exists a subharmonic (see Corollary 4.1.4.2) and hence entire (see Corollary 5.3.1.5) function f such that Fr[f ] = Λ. By Theorem 5.4.2.1, (5.5.1.1) holds.  For some M it is possible to compute the supremum in (5.5.1.1) and thus to obtain explicit precise estimates of indicators in the respective class A(M). As an example, we shall present an estimate given by A.A. Gol dberg. We recall that the upper density of zeros of an entire function f ∈ A(ρ) is defined by the equality nf (r) Δ[nf ] := lim sup ρ r r→∞

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Chapter 5. Applications to Entire Functions

where nf is the distribution of zeros of the function f , and denote

− d + iφ G (e /t) K(t, φ) := − dt p

(5.5.1.2)

where a+ := max(a, 0), a− := min(a, 0). This function is piecewise continuous. Corollary 5.5.1.2 [Go(1962)] Let the distribution of zeros nf of a function f be concentrated on the positive ray, and let Δ[nf ] ≤ Δ < ∞. Then ∞ h(φ, f ) ≤ Δ

tρ K(t, φ)dt, φ ∈ [0, 2π)

(5.5.1.3)

0

and there exists a function from the same class for which equality is attained for all φ. Proof of Corollary 5.5.1.2. We exploit Theorem 5.5.1.1. The class of functions f satisfying the assumption of the corollary coincides with the class of f for which Fr[nf ] ⊂ M = {ν ∈ M(ρ) : supp ν ⊂ [0, ∞] ∧ ν(r) ≤ Δrρ }.

(5.5.1.4)

Exercise 5.5.1.1 Show this by using Corollary 3.3.2.6. Thus ∞

∞ Gp (e /t)ν(dt) ≤

iφ

iφ

Π(e , ν, p) = 0

iφ G+ p (e /t)ν(dt). 0

Integrating by parts we obtain ∞ Π(e , ν, p) ≤ − iφ

ν(t)[

d + iφ G (e /t)]− dt. dt p

0



By (5.5.1.4) we get (5.5.1.3). We write Mp (r) := max{Gp (reiφ ) : φ ∈ [0, 2π)}. In the same way one can prove

Corollary 5.5.1.3 [Go(1962), Thm. 4.1] Let distribution of zeros of the function f ∈ A(ρ) satisfy only the condition Δ[nf ] ≤ Δ < ∞. Then ∞ h(φ, f ) ≤ Δρ

tρ−1 Mp (1/t)dt, φ ∈ [0, 2π)

(5.5.1.5)

0

and there exists a function from the same class for which equality is attained for all φ.

5.5. Asymptotic extremal problems. Semiadditive integral

175

Exercise 5.5.1.2 Prove this corollary exploiting M := {ν ∈ M(ρ) : ν(r) ≤ Δrρ , ∀r > 0}. 5.5.2 To be able to obtain explicit estimates for more diverse classes of entire functions defined by a restriction on the density of zeros, Gol dberg introduced an integral with respect to a nonadditive measure and obtained estimates for indicators in terms of a one-dimensional integral (along a circumference) with respect to such a measure ([Go(1962)]. Gol dberg initially constructed the integral sum of a special form. The construction presented here is based on the Levin–Matsaev– Ostrovskii theorem (see [Go(1962), Thm. 2.10]). Fainberg (1983) developed this approach using a two-dimensional integral. This made it possible to extend significantly the set of classes of entire functions for which the estimate expressed by a nonadditive integral is precise. We shall present these results after the necessary definitions. Let δ(X) be a nonnegative monotonic function of X ⊂ C, the function being finite on bounded sets and δ(∅) = 0. For a given family of sets X := {X} we denote by N (δ, X ) the class of countable-additive measures μ defined by the relation N (δ, X ) := {μ : μ(X) ≤ δ(X), X ∈ X }. For a Borel function f ≥ 0 we define the quantity 

 (X )

f dδ := sup

 f dμ : μ ∈ N (δ, X ) ,

called an (X )-integral with respect to a nonnegative measure δ. For a Borel set E ⊂ C we set   f dδ := (X ) f IE dδ, (X ) E

where IE is an indicator of the set E, i.e.,  IE (z) :=

1, if z ∈ E; 0 if z ∈ / E.

This integral possesses a number of natural properties: it is monotonic with respect to f and δ and the family X , positively homogeneous and semi-additive with respect to the function f and δ. If δ is a measure, if X is a Borel ring, and if f is a measurable function, then (X )-integral coincides with the Lebesgue-Stieltjes integral. Exercise 5.5.2.1 Check these properties.

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Chapter 5. Applications to Entire Functions

Let δ(Θ) be a nonadditive measure on the unit circle T, defined initially on the family of all open sets Θ ⊂ T. It can be naturally extended to all closed sets ΘF using the equality δ(ΘF ) := inf{δ(Θ) : Θ ⊃ ΘF }. Let χΘ be a set of open sets containing the set T. We write Dr,Θ := {z = teiθ : 0 < t < r, eiθ ∈ Θ}, χz := {Dr,Θ : r > 0, Θ ∈ χΘ }. The subscripts Θ and z at χ indicate that the families under consideration are located either on T or on the plane, respectively. Let us define a nonadditive measure δz on χz by the equalities 

δz (Dr,Θ ) := rρ δ(Θ), Dr,Θ ∈ χz .

iθ Now the integral (χz ) G+ p (e /ζ)dδz is defined. Recall that the classical angular upper density of zeros of an entire function f ∈ A(ρ) is defined by the equality (compare (3.3.2.7)) cl

Δ [nf , Θ] := lim sup nf (Dr,Θ )r−ρ . r→∞

Consider the class of entire functions Acl (δ, χΘ ) defined by the equality cl

Acl (δ, χΘ ) := {f : Δ [nf , Θ] ≤ δ(θ), ∀Θ ∈ χΘ }

(5.5.2.1)

for a given non-additive measure δ(Θ) and a family χΘ . Theorem 5.5.2.1 [Fa] Let δ(Θ) satisfy the condition δ(Θ) = δ(Θ), ∀Θ ∈ χΘ

(5.5.2.2)

(the dash means the closure of a set). Then  iθ h(φ, f ) ≤ (χz ) G+ p (e /ζ)dδz .

(5.5.2.3)

There exists a function f ∈ Acl (δ, χΘ ) such that equality in (5.5.2.7) is attained for all φ ∈ [0, 2π) simultaneously. cl

Proof. Let us note the following: If we replace in this theorem Δ [nf , Θ] with its D counterpart Δ(CoΘ (I1 )) (see Theorem 3.3.1.2) and consider the corresponding class of entire functions A(δ, χΘ ), the assertion of the theorem holds without conditions (5.1.5.6). You should only apply Theorem 5.5.1.1 with the corresponding M. The condition (5.5.2.7) is exploited only for replacing “D ” quantities by the classic ones using results of Sections 3.3.2.  Exercise 5.5.2.2 Prove this theorem in detail. It is also worth noting that every family χΘ can be replaced by a family χΘ that is dense in χΘ (see 3.2.2) and such that for χΘ (5.5.2.6) already holds (see Theorem 3.3.2.3).

5.6 Entire functions of completely regular growth

177

5.6 Entire functions of completely regular growth. Levin-Pfluger Theorem. Balashov’s theory 5.6.1 The most famous definition of a function of completely regular growth (CRGfunction) is the following: A function f ∈ A(ρ(r)) is a function of completely regular growth, if the limit lim r−ρ(r) log |f (z)|, r := |z| z→∞

exists when z → ∞ uniformly outside some C01 -set (see Section 5.2.1.) Actually, it is equivalent to all other definitions of the functions of completely regular growth in the plane (compare [Le, Ch. III], [Pf(1938)], [Pf(1939)]). By A.A. Gol dberg ([Go(1967)]) this definition was reduced to the following: A function f ∈ A(ρ(r)) is a function of completely regular growth, if hf (φ) = hf (φ), ∀φ ∈ [0, 2π). Because of the formulae (3.2.1.1), (3.2.1.2) (see also Section 3.2.7) we have the following Theorem 5.6.1.1 A function f ∈ A(ρ(r)) is a function of completely regular growth (CRG-function) iff Fr[f ] consists of only one subharmonic function h(z). Because of (3.2.1.11) the function h(z) has the form h(z) = rρ h(eiφ ).

(5.6.1.1)

The function h(φ) := h(eiφ ) is ρ-trigonometrically convex and it was studied in Sections 3.2.3, 3.2.4, 3.2.5. 5.6.2 The initial definition of regular zero distribution [Le, Ch. II, § 1] is the following: Let n be a zero distribution (divisor, or mass distribution) of convergence exponent ρ1 := ρ[n] (see Section 2.8.3), and let ρ1 > [ρ1 ]. Let ρ1 (r) → ρ1 be a proper proximate order of n(r) (see Theorem 2.8.1.2). It means that n ∈ M(ρ(r)), ρ(r) → ρ1 (see Section 3.1.3). The initial definition of regular zero distribution for ρ1 being non-integer is: A zero distribution n is regular if the limit lim

r→∞

n(Co(α,β) (It )) := Δ((α, β)) tρ1 (t)

1

exists for all α > β except perhaps for a countable set on the circle. 1 For

the definition of Co(α,β) (It ), see Exercise 3.3.1.5.

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Chapter 5. Applications to Entire Functions

By using results of Section 3.3, one can prove Theorem 5.6.2.1 The zero distribution n is regular iff Fr[n] consists of only one measure νreg . Exercise 5.6.2.1 Prove this exploiting Theorems 3.3.3.1 and 3.3.2.4. Recall that for f ∈ A(ρ(r)), ρ(r) → ρ we have nf ∈ M(ρ(r)), ρ(r) → ρ (see Theorem 2.9.3.2). Now we can formulate Theorem 5.6.2.2 (Levin-Pfluger) [Le, Ch. II, Ch. III] An entire function f ∈ A(ρ(r)), ρ(r) → ρ of non-integer order ρ is of completely regular growth function iff its zero distribution is regular. After Theorems 5.6.1.1, 5.6.2.1 this theorem is a direct corollary of Theorem 3.1.5.1. 5.6.3 Consider now the case of integer ρ. In general, this case differs from the case of non-integer ρ. For example, Theorem 2.9.4.2 (Brelot-Lindel¨ of) implies that (f ∈ A(ρ(r)), ρ(r) → ρ) ⇐⇒ nf ∈ M(ρ(r)), ρ(r) → ρ iff the family of polynomials (2.9.4.4a) is compact. To describe the regularity of zero distribution for the case of integer ρ we assume that the limit (5.6.3.1) lim δR (z, ν, ρ) := [δ∞ z ρ ] R→∞

exists, where ⎡ ⎢ δ∞ := lim ⎣



R→∞

|ζ|
|ζ|−ρ cos arg ζn(dζ) + i



⎤ ⎥ |ζ|−ρ sin arg ζn(dζ)⎦ .

|ζ|
Now a zero distribution n ∈ M(ρ(r)), ρ(r) → ρ with integer ρ is called regular if Fr[n] consists of only one measure νreg as in Theorem 5.6.2.1 and the limit (5.6.3.1) exists. Under this definition Theorem 5.6.2.2 still holds, because the set (H, Fr)[log |f |] from Theorem 3.1.5.2 consists of only one element ([δ∞ z ρ ], νreg ). Note also Proposition 5.6.3.1 The measure νreg has the form νreg (drdφ) = ρrρ dr ⊗ Δ(dφ) where Δ is a measure of bounded variation on the unit circle. This assertion is a corollary of invariance of Fr[n], Theorem 3.1.3.3, frm3).

5.6. Entire functions of completely regular growth

179

5.6.4 In the papers [Bal(1973), Bal(1976)] functions of completely regular growth along curves of regular rotation were considered. A curve of regular rotation is a curve that is described by the equation z = tei(γ(t) log t+φ) , 0 < t < ∞ for a fixed φ. If γ(t) ≡ γ, then this curve is a logarithmic spiral. In the general case γ(t) is a differentiable function such that γ(t) → γ, tγ  (t) → 0, t → ∞. To describe this theory in terms of limit sets we consider the transformation Pt = teiγ(t) log t , ut (z) = u(Pt z)t−ρ(t) . The following theorem is similar to Theorem 3.1.2.1. Proposition 5.6.4.1 (Existence of spiral Limit Set) The following holds: esls 1) ut ∈ SH(ρ(r)) for all t ∈ (0, ∞); esls 2) the family {ut } is precompact at infinity. The set of all limits D − lim utj does not depend on γ(t) but only on the j→∞

constant γ since

lim (γ(t) − γ) log t = lim tγ  (t) = 0.

t→∞

t→∞

So it is the same as that for Pt = teiγ log t , i.e., the case that was already considered in the general theory. In particular (3.2.1.8) for this case has the form z 0 (z) = ei(−γ log r+φ) .

(5.6.4.1)

Hence, from Theorem 3.2.1.2 the indicator (see (3.2.1.1)) has the form h(reiφ ) = rρ h(−γ log r + φ), z = reiφ , where h(φ) is a ρ-trigonometrically convex 2π-periodic function (see Section 3.2.3). All other assertions of Levin-Pfluger theory can be obtained analogously from other general assertions as it was done in the previous sections. Theorem 3.1.6.1 connects limit sets for every γ. Exercise 5.6.4.1 Formulate and prove Balashov’s analogy of the Levin-Pluger Theorem 5.6.2.2 and Theorem 3.1.6.1 for m = 2. For other generalizations of the Levin-Pfluger theory see [AD] and [Az(2007)].

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Chapter 5. Applications to Entire Functions

5.7 General characteristics of growth of entire functions 5.7.1 A functional F (u) acting in the unit circle and defined on subharmonic functions u ∈ SH(ρ(r)) is called a growth characteristic if the following conditions are fulfilled: 1. continuity: (5.7.1.1) F (uj ) → F (u), if uj → u uniformly on compacts (of course, for continuous functions u) or if uj ↓ u; 2. positive homogeneity: F (cu) = cF (r, u);

(5.7.1.2)

for every constant c > 0. Here we shall list some widely used functionals that satisfy these conditions: Hφ (u) := u(eiφ ); T (u) =

1 2π

(5.7.1.3)

2π u+ (eiφ )dφ;

(5.7.1.4)

0

Mα (u) := max{u(eiφ ) : |φ| ≤ α}; M (u) := Mπ (u);

(5.7.1.5) (5.7.1.6)

β Iαβ (u) :=

u(eiφ )dφ;

(5.7.1.7)

u(eiφ )g(φ)dφ, g ∈ L1 [0, 2π].

(5.7.1.8)

α

2π I(u, g) := 0

Exercise 5.7.1.1 Check properties 1 and 2 for these functionals. Let α(t) and α (ζ) be the “hats” defined by the equalities (2.3.1.1)–(2.3.1.3) and let R u be defined by (2.3.1.4). This averaging has the following properties. Proposition 5.7.1.1 1. if u is subharmonic, then R u is subharmonic; 2. R u ↓ u as ↓ 0 for every subharmonic function; 3. if uj → u in D and uj , u are locally summable functions, R uj → R u uniformly on every compact set.

5.7. General characteristics of growth of entire functions

181

Exercise 5.7.1.2 Prove this using Theorem 2.3.4.5, 2.6.2.3. Now we can define the asymptotic characteristics of growth of entire function f ∈ A(ρ(r)) : F [f ] := lim lim sup F (R ut (•)),

(5.7.1.9)

F [f ] := lim lim inf F (R ut (•)),

(5.7.1.10)

→0

t→∞

→0 t→∞

where u = log |f | and (•)t is defined by (3.1.2.1). Proposition 5.7.1.2 For F (u) defined by (5.7.1.3) F [f ] = hf (φ); F [f ] = hf (φ). For other functionals from the list (5.7.1.4)–(5.7.1.8) one may replace R u by u and omit lim . →0

Exercise 5.7.1.3 Prove this. The following assertion connects the asymptotic growth characteristics with limit sets. Theorem 5.7.1.3 The relations F [f ] = sup{F (v) : v ∈ Fr[f ]}, F [f ] = inf{F (v) : v ∈ Fr[f ]} are true. Proof. Let v ∈ Fr[f ] and utj → v in D . Then R utj → R v uniformly on every compact set. Hence lim F (R utj ) = F (R v). tj →∞

Passing to the limit as → 0 we obtain lim lim F (R utj ) = F (v).

→0 tj →∞

Choosing a sequence that corresponds to lim sup or lim inf we obtain the assertion of the theorem.  Applying this theorem to the functional (5.7.1.3) we obtain the RHS’s of (3.2.1.1), (3.2.1.2) and hence another definition for the indicator and lower indicator. 5.7.2 A family of growth characteristics χA := {Fα (r, •) : α ∈ A} is called total if the equation Fα (v1 ) = Fα (v2 ), ∀r > 0, α ∈ A (5.7.2.1) implies v1 ≡ v2 for v1 , v2 ∈ U [ρ] (see 3.1.2.4).

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Chapter 5. Applications to Entire Functions

Here are some examples of the total families: χH := {Hφ (u(eiφ ) : φ ∈ [0, 2π)}; χI := {Iα,β (u) : α, β ∈ [0, 2π)};

(5.7.2.2) (5.7.2.3)

χF o := {ck (u) = I(u, gk ) : k ∈ Z};

(5.7.2.4)

g0 := 1, gk := cos kφ; g−k = sin kφ, k ∈ N.

(5.7.2.5)

where It is easy to deduce from Theorem 5.6.1.1 Theorem 5.7.2.1 Let a family {Fα (•) : α ∈ A} be a total family of characteristics. An entire function f is a CRG-function iff F α [f ] = F α [f ].

(5.7.2.6)

Exercise 5.7.2.1 Check this. 5.7.3 Let us consider a total family of characteristics of the form χΨ := {I(u, ψ) : ψ ∈ Ψ},

(5.7.3.1)

where Ψ is a set which is complete in L1 [0, 2π]. For instance, such are the families χI and χF o . Theorem 5.7.3.1 [Po(1985)] Let f ∈ A(ρ(r)). The following assertions are equivalent: a) F [f g] = F [f ] + F [g], ∀F ∈ χΨ , b) F[f g] = F [f ] + F [g], ∀F ∈ χΨ , for all entire functions g ∈ A(ρ(r)). c) f is a GRG-function. Proof of sufficiency of assertion c). Let us prove c) =⇒ a) and c) =⇒ b). Using Theorem 5.7.1.3 we obtain for every characteristic F F [f g] = sup{F (w) : w ∈ Fr[f g]}.

(5.7.3.2)

Because of Theorem 3.1.2.4 fru1), Fr[f g] ⊂ Fr[f ] + Fr[g]. Since f is a CRG-function, Fr[f ] consists of only one subharmonic function vreg (see Theorem 5.6.1.1) and it is easy to check that in this case we have equality Fr[f g] = vreg + Fr[g]. Exercise 5.7.3.1 Check this.

5.7. General characteristics of growth of entire functions

183

Since F (vreg + vg ) = F (vreg ) + F (vg ), we obtain F[f g] = F (vreg ) + sup{F (vg ) : vg ∈ Fr[g]} = F [f ] + F [g]. So c) =⇒ a) was proved. In the same way one can prove c) =⇒ b).



Exercise 5.7.3.2 Prove this. In the proof of sufficiency for c) of a) and b) we can suppose that ψ belong to the space D(T) of infinitely differentiable functions on the unit circle T because D(T) is complete in L1 [0, 2π]. We prove now sufficiency of b) in Theorem 5.7.3.1. We recall that (see (3.1.2.4a)) v[t] (z) = v(tz)t−ρ , v ∈ U [ρ] to distinguish it from (•)t that we define as ut (z) = u(tz)t−ρ(t) , u ∈ SH(ρ(r)). The main constructive element for the proof sufficiency of b) in Theorem 5.7.3.1 is Lemma 5.7.3.2 Let ψ 0 ∈ D(S). There exists v ∈ U [ρ] with the following properties: D − lim v[t] = D − lim v[t] = v˜, t→∞

t→0 i•

v[t] (ei• ), ψ 0  > v(e ), ψ 0  for t ∈ (0, ∞), t = 1, ˜ v (e ), ψ  > v(e ), ψ . i•

0

i•

0

(5.7.3.3) (5.7.3.4) (5.7.3.5)

Proof of Lemma 5.7.3.2. Let ψ 0 be represented by Fourier series ∞

ψ 0 (φ) =

a0  (an cos nθ + bn sin nθ) + 2 n=1

Since ψ 0 ≡ 0 there exists ak = 0 or bk = 0. Suppose there exists ak = 0. In the proof we will consider three cases: 1. k = 0; 2. k =  0 ∧ k ≤ p; 3. k ≥ p + 1. The number ρ is supposed non-integer and p = [ρ]. Consider the case a0 = 0, a0 > 0. Set ψ(x) := log(−e−α|x| + C), α > 0, C > 1, v(z) := |z|ρ eψ(log |z|) = exp(ρ log r + ψ(log r)).

(5.7.3.6)

184

Chapter 5. Applications to Entire Functions

Applying the Laplace operator, we obtain: ∂2 1 ∂ ∂ r r v(r) = e−2x 2 eρx+ψ(x) (5.7.3.7) 2 r ∂r ∂r / ∂x  0 = exp((ρ − 2)x + ψ(x)) (ρ + ψ (x))2 + ψ  (x) , x = log r.

Δv =

Since ψ  (x) = α sgn x exp(−α|x|) → 0, ψ  (x) = −α2 sgn x exp(−α|x|) → 0 as x → ±∞, it is possible to choose α such that the expression (5.7.3.7) is positive. So v(z) is subharmonic. It is easy to check that all the assertions of the lemma are satisfied and v˜(z) = b|z|ρ where b(> 0) is a constant. Exercise 5.7.3.3 Check this. If a0 = −|a0 | < 0, consider the function  log |z|, |z| ≥ 1, 0 v (z) := 0, |z| < 1; it is subharmonic and

0 v[t] (ei• ), ψ 0  = a0 t−ρ log+ t.

(5.7.3.8)

−1

Since the RHS of (5.7.3.8) is minimized for t0 = eρ , the function v(z) := v[t0 0 −1 ] (z) satisfies the assertions of the lemma with v˜ = lim v[t] = 0 t→0,∞

Now let a0 = 0, ak = 0, 0 < k < p. We will search for a function v of the form 2π Gp (rei(φ−θ) )(1 − sgn ak cos kθ)dθ.

iφ

v(re ) :=

(5.7.3.9)

0

This is the convolution Gp (rei• ) ∗ g of the primary kernel (see Section 2.9.1) Gp (z) = log |1 − z| + 

p 

z n /n

n=1

with a positive function g(θ) := (1 − sgn ak cos kθ) on the circle. So it is subharmonic. Recall that the cos-Fourier coefficients of the function Gp (reiθ ) are (see Exercise 2.3.7.2).  m = 0, 1, . . . , p ˆ p (m, r) = 0, G if r ≤ 1, (5.7.3.10) m (1/m)r , m = p + 1, . . .

5.7. General characteristics of growth of entire functions

and

⎧ ⎪ ⎨ log r, ˆ p (m, r) = 1 (rm − r−m ), G m ⎪ ⎩ (1/m)rm ,

m=0 m = 1, . . . , p if r ≥ 1. m = p + 1, . . .

185

(5.7.3.11)

All the sin-Fourier coefficients are equal to zero. The Fourier coefficients of the function g are 1 and − sgn ak . Using well-known properties of Fourier coefficients, we obtain for 0 < k ≤ p,  0, t ≤ 1, vˆ[t] (0) = log t t ≥ 1, tρ ,  0, t ≤ 1, vˆ[t] (k) = 1 tk −t−k − k tρ sgn ak t ≥ 1,  0, t ≤ 1, v[t] (ei• ), ψ 0  = k−ρ −k−ρ −t )|ak | t ≥ 1. −1/k(t The function t → vt (ei• ), ψ 0  tends to zero when t → 0, ∞ and has its only minimum at the point  1/k ρ+k . t0 = ρ−k Thus v[(t0 )−1 ] satisfies the conditions of the lemma with v˜ = 0. For k ≥ p + 1 we should take the same g and then  −(1/k)tk−ρ |ak |, t ≤ 1, i• 0 vt (e ), ψ  = −k−ρ −(1/k)t |ak | t ≥ 1. So the corresponding function t → vt (ei• ), ψ 0  obtains its minimum at the point t0 = 1 and the function v satisfies the assertions of the lemma with v˜ = 0.  Exercise 5.7.3.4 Prove the lemma for the case bk = 0. Lemma 5.7.3.3 Let v ∈ U [ρ] with the condition D − lim v[t] = D − lim v[t] = v˜ t→∞

t→0

fulfilled, and let u ∈ SH(ρ(r)) with some v 0 ∈ Fr[u]. Then there exists w0 ∈ SH(ρ(r)) such that (5.7.3.12) Fr[w0 ] = {v[t] : t ∈ (0, ∞)} ∪ v˜ and the following condition holds: 1. If the sequence lim wt0n = v[t] for some t ∈ (0, ∞) and the sequence utn tn →∞

0 converges in D as tn → ∞, then lim utn = v[t] . n→∞

For the proof see Corollary 4.4.1.4.

186

Chapter 5. Applications to Entire Functions

Lemma 5.7.3.4 Let w ∈ SH(ρ(r)), ψ ∈ D(S). Then the following holds: lim inf w, ψ = min v, ψ, t→∞

v∈Fr w

lim supw, ψ = max v, ψ. v∈Fr w

t→∞

Exercise 5.7.3.5 Prove this exploiting completeness of Fr. Proof of sufficiency of b) in Theorem 5.7.3.1. In assumption b) we should prove that f is a CRG-function, i.e., by Theorem 5.6.1.1 its Fr[f ] consists of only one function. Since log |f | ∈ SH(ρ(r)) and because of Theorem 5.3.1.4 (Approximation) it is enough to prove the corresponding theorem for subharmonic functions. Suppose (5.7.3.13) F [u + w] = F [u] + F [w], ∀F ∈ χΨ for all w ∈ SH(ρ(r). We exploit Lemma 5.7.3.4 and write (5.7.3.13) in the form: min v∈Fr[u+v]

v, ψ = min v, ψ + min v, ψ, ∀ψ ∈ Ψ. v∈Fru

v∈Frw

Suppose the contrary, i.e., u is not a CRG-function and Fru does not consist of only one vmin ∈ U [ρ]. Then there exists v 0 = vmin . The family χΨ is total; therefore there exists ψ 0 ∈ Ψ such that v 0 , ψ 0  = vmin , ψ 0  and hence v 0 , ψ 0  > vmin , ψ 0 .

(5.7.3.14)

Using Lemma 5.7.3.2, construct for the function ψ 0 a function v ∈ U [ρ] satisfying the conditions (5.7.3.3), (5.7.3.4) and (5.7.3.5). Apply Lemma 5.7.3.3 to construct a function w0 satisfying (5.7.3.12) and the condition 1. Under conditions of the theorem, min

ω∈Fr(u+w 0 )

ω, ψ 0  = min ω, ψ 0  + ω∈Fr(u)

min ω, ψ 0 .

ω∈Fr(w 0 )

(5.7.3.15)

Let γ ∈ Fr(u + w0 ) be the function on which the minimum of LRH in (5.7.3.15) is attained. Using (5.7.3.4), (5.7.3.5) and (5.7.3.12), we can rewrite (5.7.3.15) in the form (5.7.3.16) γ, ψ 0  = min ω, ψ 0  + v, ψ 0 . ω∈Fru



Since γ ∈ Fr(u + w ), γ = D − lim (u + w0 )tn . Passing to subsequences, we can 0

n→∞

suppose that the sequences {utn } and {wt0n } have limits. Since Frw0 has the form (5.7.3.12), there are two possible cases : wt0n → v[t] , t ∈ (0, ∞) and wt0n → v˜. 0 Consider the first case. Because of condition 1 from Lemma 5.7.3.3, utn → v[t] 0 and γ = v[t] + v[t] . Substituting this in (5.7.3.15), we obtain 0 , ψ 0  − min ω, ψ 0  = v, ψ 0  − v[t] , ψ 0 . v[t] ω∈Fru

5.7. General characteristics of growth of entire functions

187

This equality leads to a contradiction because for t = 1 it contradicts (5.7.3.14) and for t = 1 it contradicts (5.7.3.4). Consider the second case, when wt0n → v˜. Denote v 2 = lim utn and rewrite n→∞

(5.7.3.15) in the form v 2 , ψ 0  − min ω, ψ 0  = v, ψ 0  − ˜ v , ψ 0 . ω∈Fru



The last equality contradicts (5.7.3.5).

Sufficiency of condition a) of Theorem 5.7.3.1 can be proved using the Lemmas 5.7.3.3, 5.7.3.4 and a variation of Lemma 5.7.3.2. Lemma 5.7.3.2 Let ψ 0 ∈ D(S). There exists v ∈ U [ρ] with the following properties: D − lim v[t] = D − lim v[t] = v˜, t→0 i•

t→∞

v[t] (ei• ), ψ 0  < v(e ), ψ 0  for t ∈ (0, ∞), t = 1, ˜ v (ei• ), ψ 0  < v(ei• ), ψ 0 .

(5.7.3.3 ) (5.7.3.4 ) (5.7.3.5 )

Exercise 5.7.3.6 Prove this lemma and sufficiency of a) in Theorem 5.7.3.1. 5.7.4 In this section we consider the question of summing the asymptotic characteristics connected with the functional (5.7.1.3), i.e., indicator and lower indicator. Recall that f ∈ A(ρ(r)) is completely regular on the ray {arg z = φ} (f ∈ Areg,φ ) if hf (φ) = hf (φ).

(5.7.4.0)

We are going to prove the following assertions: Theorem 5.7.4.1 Let f ∈ Areg,φ . Then for every g ∈ A(ρ(r)), hf g (φ) = hf (φ) + hg (φ),

(5.7.4.1)

hf g (φ) = hf (φ) + hg (φ).

(5.7.4.2)

Theorem 5.7.4.2 Suppose the equality (5.7.4.1) holds for every g ∈ A(ρ(r)). Then f ∈ Areg,φ . Let us note that the assertion of Theorem 5.7.4.2 holds also if the equality (5.7.4.1) fulfilled for some sequence φn → φ, because the indicator is a continuous function (see Section 3.2.5). So if the equality (5.7.4.1) holds for the set Φ of φ that is dense in [0, 2π) (or the set eiΦ := {eiφ : φ ∈ Φ}

(5.7.4.3)

is dense on the unit circle), then f ∈ Areg,φ for all φ, i.e., f is a CRG-function.

188

Chapter 5. Applications to Entire Functions

On the other hand, the following assertion holds Theorem 5.7.4.3 If the set Θ of θ is not dense in [0, 2π), there exists f ∈ Areg,θ , θ ∈ Θ that is not a CRG-function. The situation with a lower indicator is analogous, but in another topology. A set E is called non-rarefied at a point z0 if for every function v subharmonic in a neighborhood of z0 the following holds: v(z0 ) =

lim sup

v(z) = lim sup v(z).

z∈E,z→z0 ,z=z0

z∈E,z→z0

A set is rarefied if it is not non-rarefied. Note that if hf (φ) = −∞, then hf g (φ) = −∞ for every g ∈ A(ρ(r)). It is obvious that f ∈ / Areg,φ . The next theorems were proved in [GPS]. Theorem 5.7.4.4 Let (5.7.4.2) be fulfilled for ψ ∈ E for all g ∈ A(ρ(r)) and eiE be non-rarefied at the point eiφ . Then f ∈ Areg,φ . Theorem 5.7.4.5 Let E0 be a set such that eiE0 is rarefied at all points of the unit circle. Then there exists f ∈ A(ρ(r)) for which (5.7.4.2) is fulfilled for all φ ∈ E0 and all g ∈ A(ρ(r)), but f ∈ / Areg,φ for all φ and hf (φ) > −∞, ∀φ. Let us note that E0 can be dense in [0, 2π) and E from Theorem 5.7.4.4 can even be of zero measure. The proof of Theorems 5.7.4.4 and 5.7.4.5 is based on the following assertion that gives a criterion for (5.7.4.2) in terms of limit sets Fr[f ]. Theorem 5.7.4.6 Let f ∈ A(ρ(r)) and hf (φ) > −∞. The condition (5.7.4.2) holds for every g ∈ A(ρ(r)), such that hg (φ) > −∞ iff lim inf v(teiφ ) = hf (φ) t→1

(5.7.4.4)

for all v ∈ Fr[f ]. An analogous criterion holds for (5.7.4.1). Theorem 5.7.4.7 Let f ∈ A(ρ(r)). Then (5.7.4.1) holds for every g ∈ A(ρ(r)), iff lim sup v(teiφ ) = hf (φ), t→1

for all v ∈ Fr[f ]. Corollary 5.7.4.8 The equality (5.7.4.5) implies f ∈ Areg,φ .

(5.7.4.5)

5.7. General characteristics of growth of entire functions

189

Actually, for every v ∈ Fr[f ] we have, using semicontinuity of subharmonic functions and the definition (3.2.1.1) of the indicator, hf (φ) = lim sup v(teiφ ) ≤ v(eiφ ) ≤ hf (φ) t→1

for all v ∈ Fr[f ]. So Fr[f ] consists of functions v that coincide at the point eiφ and hence on the ray {reiφ : r ∈ (0, ∞)}. Note also that the set eiE for which (5.7.4.1) holds is closed and Theorem 5.7.4.4 means that the set where (5.7.4.2) holds is thinly closed, i.e., closed in thin topology (see [Br, § 6]). Therefore if eiφ0 is a limit point of eiE in the euclidian (respectively, thin) topology, then (5.7.4.1) ((5.7.4.2), respectively) is also a sufficient condition for completely regular growth at φ0 . 5.7.5 The main constructive element for proving Theorem 5.7.4.6 is Lemma 5.7.5.1 Let > 0, t0 > 0 and φ0 ∈ [0, 2π) be fixed. Then there exists v ∈ U [ρ] with the following properties: D − lim v[t] = D − lim v[t] = 0,

(5.7.5.1)

v(eiφ0 ) < v[t] (eiφ0 ), t ∈ (0, 1) ∪ (1, ∞),

(5.7.5.2)

t→∞

t→0

−∞ < v(e

iφ0

) < − ,

(5.7.5.3)

and the inequality v[t] (eiφ0 ) − v(eiφ0 ) ≤ /2

(5.7.5.4)

t ∈ [1/t0 , t0 ].

(5.7.5.5)

implies The last condition means that the function ψ(t) := v[t] (eiφ0 ) can be less than ψ(1) + /2 only in a neighborhood of t = 1. Proof. Set w(z) := max(log |1 − ze−iφ0 |, −N ) + 

p  1 (ze−iφ0 )n , n n=1

N > 0, p = [ρ].

(5.7.5.6)

It is obvious that w is subharmonic, with masses νw concentrated in a neighborhood of the point eiφ0 . Thus νw ∈ M[ρ] (see (3.1.3.4)) and D − lim (νw )[t] = D − lim (νw )[t] = 0. t→0

t→∞

Hence (see Theorem 3.1.4.2) w ∈ U [ρ], and (see (3.1.5.0)) D − lim w[t] = D − lim w[t] = 0. t→0

t→∞

(5.7.5.7)

190

Chapter 5. Applications to Entire Functions

Let us capitalize on the behavior of w[t] on the ray {arg z = φ0 }: w[t] (e

iφ0

 p  1 −ρ n ) := ψ(t) = max(log |1 − t|, −N ) +  t t . n n=1

(5.7.5.8)

It is possible to prove directly the following properties of ψ(t). i) outside interval (1−e−N , 1+e−N ), ψ(t) = Gp (t)t−ρ ; where Gp is the Primary Kernel (see Section 2.9.1) and inside this interval the first summand is −N ; ii) ψ(t) > 0 for t > t1 where t1 is a zero of the equation Gp (t) = 0, ψ(t) decreases monotonically on the interval (0, 1 − e−N ) and increases monotonically on the interval (1 − e−N , t1 ). Exercise 5.7.5.1 Prove this. Now set t2 := 1 − e−N and v(z) := Dwt2 (z), where D is a constant. This function satisfies the conditions (5.7.5.1) and (5.7.5.2) of the lemma and v[t] (eiφ0 ) has only one negative minimum for t = 1. Thus it is possible to take D sufficiently  large to satisfy the conditions (5.7.5.3) and (5.7.5.4) for fixed and t0 . Exercise 5.7.5.2 Prove this in detail. In the proof of Theorem 5.7.4.6 we also use Lemma 5.7.3.3. We can prove all the assertions for subharmonic functions from SH(ρ(r)). Proof of Theorem 5.7.4.6. Necessity. We should prove that if the equality h(eiφ0 , u + w) = h(eiφ0 , u) + h(eiφ0 , w)

(5.7.5.9)

holds for a fixed u ∈ SH(ρ(r)), φ0 and every w ∈ SH(ρ(r)), then lim inf v(teiφ ) = h(eiφ , u) t→1

(5.7.5.10)

for all v ∈ Fru. Assume that h(eiφ , u) > −∞ and h(eiφ0 , w) > −∞. Suppose the contrary, i.e., there exists v 0 ∈ Fru such that lim inf v 0 (teiφ0 ) > h(eiφ0 , u). t→1

(5.7.5.11)

The inequality (5.7.5.11) implies that there exists > 0 and t0 > 0 such that for every t ∈ [1/t0 , t0 ] the inequality v 0 (teiφ0 ) > h(eiφ0 , u) +

(5.7.5.11a)

holds. Let us construct by Lemma 5.7.5.1 for these , t0 , φ0 a function v and by Lemma 5.7.3.3 for the functions u, v 0 and the already found v a function w0 . Let us show that for w0 the equality (5.7.5.9) does not hold.

5.7. General characteristics of growth of entire functions

191

Compute h(eiφ , w0 ). From (3.2.1.2) h(eiφ0 , w0 ) = min{0, inf{v[t] (eiφ0 : t ∈ (0, ∞)}}. The inequalities (5.7.5.3) imply that 0 can be omitted and (5.7.5.2) implies that the infimum is attained at t = 1, i.e., h(eiφ0 , w0 ) = v(eiφ0 ).

(5.7.5.12)

Find v  ∈ Fr(u + w0 ) such that h(eiφ0 , u + w0 ) > v  (eiφ0 ) − /3. Let tn → ∞ and (u + w0 )tn → v  in D . Passing to subsequences we can assume that utn and wt0n also converge. Consider two cases. The first, when D − lim wt0n = v[t] , t ∈ (0, ∞).

(5.7.5.13)

0 0 and hence v  = lim(w0 + u)tn = v[t] + v[t] . If By Lemma 5.7.3.3 lim utn = v[t] t∈ / [1/t0 , t0 ], then by (5.7.5.4)

v[t] (eiφ0 ) > v(eiφ0 ) + /2 = h(eiφ0 , w0 ) + /2.

(5.7.5.14)

In this case we have 0 − /3. h(eiφ0 , u + w0 ) ≥ v  (eiφ0 ) − /3 ≥ v[t] + v[t]

(5.7.5.14a)

Using (5.7.5.14a), we obtain h(eiφ0 , u + w0 ) ≥ h(eiφ0 , w0 ) + h(eiφ0 , u) + /6.

(5.7.5.15)

If t ∈ [1/t0 , t0 ], then from (5.7.5.11) we have h(eiφ0 , u + w0 ) ≥ h(eiφ0 , w0 ) + h(eiφ0 , u) + 2 /3.

(5.7.5.16)

So the case (5.7.5.13) is settled. Let D − lim wt0n = 0. In this case we have h(eiφ0 , u + w0 ) ≥ v  (eiφ0 ) − /3 ≥ h(eiφ0 , u) − + − /3 = h(eiφ0 , u) − + 2 /3. Using (5.7.5.12) and (5.7.5.3) we obtain h(eiφ0 , u + w0 ) ≥ h(eiφ0 , u) + h(eiφ0 , w0 ) + 2 /3. So we proved in any case that (5.7.5.9) does not hold if (5.7.5.10) does not hold. Let us prove sufficiency in Theorem 5.7.4.6. We prove it for subharmonic functions. Let u ∈ SH(ρ(r)) and for every v ∈ Fru (5.7.5.10) holds. Let us show that for all w ∈ SH(ρ(r)) (5.7.5.9) holds. It is sufficient to prove that h(eiφ0 , u + w) ≤ h(eiφ0 , u) + h(eiφ0 , w)

(5.7.5.17)

192

Chapter 5. Applications to Entire Functions

holds since the inverse inequality holds for every w ∈ SH(ρ(r)) (see (3.2.1.5)). Let us begin by noting that for every v 2 ∈ Frw there exist v ∈ Fr(u + w) and v 1 ∈ Fru such that (5.7.5.18) v = v1 + v2 . Indeed, let tn → ∞ be a sequence such that wtn → v 2 . We can suppose, in choosing a subsequence, that utn → v 1 and (u + w)tn → v. Then (5.7.5.18) holds. Let be arbitrarily small. Choose v 2 ∈ Frw such that v 2 (eiφ ) < h(eiφ , w) + holds. From upper semicontinuity of v 2 we have lim sup v 2 (eiφ ) ≤ h(eiφ0 , w) + .

(5.7.5.19)

t→1

Let v 1 ∈ Fru and v ∈ Fr(u + w) satisfy (5.7.5.18). Then we have 1 2 h(eiφ0 , u + w) ≤ (v 1 + v 2 )[t] (eiφ0 ) = v[t] (eiφ0 ) + v[t] (eiφ0 ), ∀t.

Hence 1 2 (eiφ0 ) + lim sup v[t] (eiφ0 ). h(eiφ0 , u + w) ≤ lim inf v[t] t→1

t→1

Using (5.7.5.10) and (5.7.5.19) we obtain h(eiφ0 , u + w) ≤ h(eiφ0 , u) + h(eiφ0 , w) + . This proves the inverse inequality and hence the equality (5.7.5.9), because is arbitrarily small.  5.7.6 Now we are going to prove Theorem 5.7.4.4. We need the following assertion from Potential Theory. Lemma 5.7.6.1 Let E be a set that is non-rarefied at the point eiφ0 . Let E  be a set in C, such that ∀eiφ ∈ E and ∀δ > 0 there exists a point z  ∈ E  on the ray {arg z = φ} such that |z  − eiφ | < δ. Then E  is also non-rarefied at the point eiφ0 . Proof. We can suppose without loss of generality that E  has no intersection with some neighborhood of zero. Denote by P (z) the map z → ei arg z . It is easy to see that for all pairs z1 , z2 ∈ E  the inequality |P (z1 ) − P (z2 )| < A|z1 − z2 | holds for some constant A. Thus the logarithmic capacity (2.5.2.5) satisfies ([La, Ch. II, § 4, it. 11, 15]). (5.7.6.1) capl (M ) < Acapl (M  ) where M  ⊂ E  , M = P (M  ). Now we exploit the following properties of nonrarefied sets. First, if E is non-rarefied at a point z0 , then there exists a compact set that is non-rarefied at z0 ([La, Ch. V, § 1, it. 5, § 3, it. 9]). Second, for a compact set K that is non-rarefied at z0 , ∞ 

n =∞ −1 log(cap l Kn ) n=1 where Kn := K ∩ {z : q n+1 ≤ |z − z0 | ≤ q n }, 0 < q < 1.

(5.7.6.2)

5.7. General characteristics of growth of entire functions

193

Using the inequality (5.7.6.1), we obtain that divergence of the series (5.7.6.2) for a compact K ⊂ E implies divergence for K  ⊂ E  where K = P (K  ), i.e., E  is non-rarefied at the point P (eiφ0 ) = eiφ0 .  Proof of Theorem 5.7.4.4. Let (φ) → 0 as φ → φ0 and let v ∈ Fru. Suppose (5.7.5.9) holds for eiφ ∈ E. By Theorem 5.7.4.6 the equality (5.7.5.10) holds. Thus ∀Δ > 0, ∃z  = z  (eiφ , Δ) such that |z  − eiφ | < Δ, arg z  = φ, v(z  ) < h(eiφ ) + (φ). Set E  :=

∞  

(5.7.6.3)

z  (eiφ , 1/n).

φ∈E n=1

By (5.7.6.3) and upper semicontinuity of h(eiφ ) we obtain lim sup z  →eiφ0 , z  ∈E 

v(z  ) ≤ h(eiφ0 ).

(5.7.6.4)

Since E  is non-rarefied, by Lemma 5.7.6.1 the upper limit of v coincides with v(eiφ0 ) and hence v(eiφ0 ) ≤ h(eiφ0 ). The inverse inequality holds always. Thus v(eiφ0 ) = h(eiφ0 ), ∀v ∈ Fru. Hence h(eiφ0 ) = h(eiφ0 ).  5.7.7 Now we are going to prove Theorem 5.7.4.5. Before this we need to describe a construction and prove some auxiliary assertions. Let Bj := {z : T j < |z| < T j+1 }, j = 0, ±1, ±2, . . . where T > 1 is a fixed number. Denote LE0 := {z : ei arg z ∈ eiE0 }. Recall that eiE0 is a set rarefied at every point of the unit circle. Let Q be the set of rational numbers on the interval (1, T ). Set SQ := {z : |z| ∈ Q}, T SQ := {zT j : z ∈ SQ }, j

Aj := LE0 ∩ T j SQ ,

j = 0, ±1, ±2, . . . .

Lemma 5.7.7.1 There exists v ∈ U [ρ] such that for z ∈ A0 and

v(z) = −∞

(5.7.7.1)

μv (e) = 0, ∀e ⊂ C \ B0 .

(5.7.7.2)

Proof. The set E is rarefied at every point, hence it is polar ([Br, Ch. 7, § 4]). Thus the set {z : |z| = r} ∩ LE0 is polar (see [Br, Ch. 3, § 2]). A countable union of polar sets is polar ([Br, Ch. 3, § 2]). Thus A0 is polar.Hence there exists a positive  measure μ concentrated on B0 for which the potential v(z) := Gp (z/ζ)μ(dζ) is equal to −∞ on A0 (see [Br, Ch. 4, § 6, Applications]). It is easy to see that μ ∈ M(ρ) and hence v ∈ U [ρ] (see Theorem 3.1.4.2). 

194

Chapter 5. Applications to Entire Functions

Lemma 5.7.7.2 There exists ω ∈ U [ρ] such that the following conditions are fulfilled: ρ (5.7.7.3) ω(z) = −∞, z ∈ A := ∪+∞ j=−∞ Aj ; ω(T z) = T ω(z). Proof. Set for every E  C \ 0, ν(E) :=

j=+∞ 

T jρ μv (T −j E ∪ B0 )

(5.7.7.4)

j=−∞

(compare Theorem 4.1.7.1). We have ν ∈ M(ρ). Set  ω(z) := Gp (z/ζ)ν(dξdη), ζ = ξ + iη. 

This ω satisfies (5.7.7.3). Exercise 5.7.7.1 Prove this using Theorem 4.1.7.1. Lemma 5.7.7.3 Let ω be a subharmonic function in C. Denote m(φ) := max{ω(reiφ ) : r ∈ [1, T ]}. Then there exists a constant C > −∞ such that m(φ) > C ∀φ. Proof. If not, there exists a sequence φn that we can assume to converge φ∞ such that m(φn ) → −∞. By upper semicontinuity of ω we have ω(z) −∞, ze−iφ∞ ∈ [1, T ]. Thus ω(z) ≡ −∞ because the capacity of the segment the plane is positive and hence it is not polar for some subharmonic function.

to = in 

Recall that for v ∈ U [ρ] (see (4.1.3.1)) C(v) := D − clos{v[t] : 0 < t < ∞}, 

(5.7.7.5)



Ω(v) := {v ∈ U [ρ] : (∃tk → ∞)(v = lim v[tk ] },

(5.7.7.6)

A(v) := {v  ∈ U [ρ] : (∃τk → 0)(v  = lim v[tk ] }.

(5.7.7.7)

k→∞

k→∞

By Theorems 4.1.3.3 and 4.2.1.2, if A(v) ∩ Ω(v) = ∅,

(5.7.7.8)

there exists u ∈ SH(ρ(r)) such that Fru = C(v).

(5.7.7.9)

5.7. General characteristics of growth of entire functions

195

Lemma 5.7.7.4 There exists v 1 ∈ U [ρ] such that the following holds: A(v 1 ) = Ω(v 1 ),

(5.7.7.10)

inf{v(eiφ ) : v ∈ C(v 1 )} = lim inf v(teiφ ) = 0, ∀v ∈ C(v 1 ), ∀eiφ ∈ eiE0 , (5.7.7.11) t→1

sup{v(eiφ ) : v ∈ C(v 1 )} = inf{v(eiφ ) : v ∈ C(v 1 )}.

(5.7.7.12.)

Proof. Let ω(z) be constructed by Lemma 5.7.7.2. Set v(z) := ω(z) + D log+ 2|z|. The condition (5.7.7.3) implies A(ω) = Ω(ω) = {ω[t] : t ∈ [1, T ]} because it is a Periodic Limit Set (see Theorem 4.1.7.1). Since (log+ 2|z|)[t] → 0, t → 0, t → ∞, the function v satisfies the condition A(v) = Ω(v) = {ω[t] : t ∈ [1, T ]}. By Theorem 2.1.7.4 for the function v 1 := v + we have + : t ∈ [1, T ]}. A(v 1 ) = Ω(v 1 ) = {ω[t]

Note that v 1 (z) = 0 for z ∈ A and since A is dense in LE0 (5.7.7.11) holds. Choosing D sufficiently large it is possible (using Lemma 5.7.7.3) to find on every ray {arg z = φ} a point zφ where v 1 (zφ ) > 0. Hence sup{v(eiφ ) : v ∈ C(v 1 )} > 0. Because of (5.7.7.11) and upper semicontinuity of inf{v(eiφ ) : v ∈ C(v 1 )} it is zero for every eiφ . Thus (5.7.7.12) holds.  Proof of Theorem 5.7.4.5. Let us construct by Theorems 4.1.3.3 and 4.2.1.2 a function u ∈ SH(ρ(r) such that Fru = C(v 1 ) where v 1 is taken from Lemma 5.7.7.4. It does not belong to Areg,φ for any φ. The equality (5.7.5.9) holds for every φ ∈ E0 because of (5.7.7.11) by Theorem 5.7.4.6.  5.7.8 The proof of Theorem 5.7.4.1 is a copy of the proof of sufficiency of assertion c) in Theorem 5.7.3.1. Exercise 5.7.8.1 Prove Theorem 5.7.4.1. Now we are going to prove Theorem 5.7.4.7 which implies (as it was shown in Corollary 5.7.4.8) Theorem 5.7.4.2.

196

Chapter 5. Applications to Entire Functions

The main constructive element of the proof of necessity is Lemma 5.7.8.1 Let > 0, t0 > 0 and φ0 ∈ [0, 2π) be fixed. Then there exists v ∈ U [ρ] with the following properties: D − lim v[t] = D − lim v[t] = 0,

(5.7.8.1)

v(eiφ0 ) > v[t] (eiφ0 ), t ∈ (0, 1) ∩ (1, ∞),

(5.7.8.2)

v[t] (eiφ0 ) − v(eiφ0 ) ≥ − /2

(5.7.8.3)

t ∈ [t0 , 1/t0 ].

(5.7.8.4)

t→∞

t→0

and the inequality implies The last condition means that the function ψ(t) := v[t] (eiφ0 ) can be more than ψ(1) − /2 only in a neighborhood of t = 1. Proof. Consider the function w(z) := log+ |z|.

(5.7.8.5)

It is subharmonic and satisfies (5.7.8.1). Since the function ψ(t) := w[t] (eiφ0 ) = t−ρ log+ t has its only strict maximum in the point tmax > 1, the function v(z) := w(z/tmax ) 

has all the properties (5.7.8.1)–(5.7.8.4).

After this lemma all the proof of Theorem 5.7.4.6 can be repeated with minimal changes. Exercise 5.7.8.2 Prove Theorem 5.7.4.7. 5.7.9 Now we are going to prove Theorem 5.7.4.3. Let us prove the following Lemma 5.7.9.1 Let Θ be a closed subset of [0, 2π). Then for every σ > 0 there exists a 2π-periodic ρ-trigonometrically convex function h(φ) such that h(φ) = σ

(5.7.9.1)

h(φ) > σ

(5.7.9.2)

for φ ∈ Θ and for φ ∈ / Θ.

5.7. General characteristics of growth of entire functions

197

Proof. We can suppose that 0 ∈ Θ, otherwise we can shift it a little. The set [0, 2π) \ Θ is open and it can be represented as the union of non-intersecting open intervals. If length of an interval is ≤ π/ρ we can construct a ρ-trigonometrical function that is equal to σ on the ends of the interval. It is greater than σ in all inner points of the interval because f (φ) ≡ σ is a strictly ρ-trigonometrical function. If the length of the interval is greater than π/ρ, for example (−l/2, l/2) with l > π/2ρ, we cover it by intersecting intervals of length less then π/ρ, construct hI (φ) as before for every interval I and set h(φ) = max hI (φ). It is obvious that I

h(φ) is greater than σ and it is ρ-trigonometrically convex.



Theorem 5.7.4.3 is a corollary of Lemma 5.7.9.1 and the following Theorem 5.7.9.2 Let h1 and h2 be two ρ-trigonometrically convex functions. Then there exists a function f ∈ A(ρ(r)) such that hf (φ) = max(h1 (φ), h2 (φ)), hf (φ) = min(h1 (φ), h2 (φ)). Proof. Consider the set U := {v(z) = crρ h1 (φ) + (1 − c)rρ h2 (φ) : 0 ≤ c ≤ 1}.

(5.7.9.3)

It consists of invariant subharmonic functions and is contained in U [ρ] and satisfies the condition of Theorem 4.1.4.1. Hence (Theorems 4.2.1.2, Corollary 5.3.1.5) there exists a function f ∈ A(ρ(r)) such that Frf = U.

(5.7.9.4)

By formulae (3.2.1.1), (3.2.1.2) we obtain the assertion of the theorem, using (5.7.9.3).  Exercise 5.7.9.1 Prove Theorem 5.7.4.3. 5.7.10 The family of characteristics {Fα , α ∈ A} is called independent if for every subset A ⊂ A (or subset in some class of subsets, for example, measurable or closed) there exists a function f = fA ∈ A(ρ(r)) such that F α [f ] = F α [f ], α ∈ A , F α [f ] = F α [f ], α ∈ A \ A . It means that for every pointed subset of characteristics there exists a function that has regular growth with respect to this subset of characteristics and is not of regular growth with respect to all other characteristics. Theorem 5.7.4.3 can be considered as an assertion of independence of the family (5.7.2.2).

198

Chapter 5. Applications to Entire Functions

Theorem 5.7.10.1 The family χF o (5.7.2.4) is independent. I.e., for every A ⊂ Z there exists f ∈ A(ρ(r) such that lim r−ρ(r)



2π

log |f (reiφ )gk (φ)dφ

r→∞

0

exists for all k ∈ A and does not exist for k ∈ Z \ A. To begin we prove Lemma 5.7.10.2 There exist two ρ-trigonometrically convex functions h1 and h2 for which  2π  2π h1 (φ)gk (φ)dφ = h2 (φ)gk (φ)dφ, k ∈ A, (5.7.10.1) 0



0



2π

2π

h1 (φ)gk (φ)dφ = 0

h2 (φ)gk (φ)dφ, k ∈ Z \ A.

(5.7.10.2)

0

Proof. Let g(φ) ∈ C 2 be a function, the Fourier coefficients of which with indices k ∈ A are equal to zero. We can represent it as a difference of ρ-trigonometrically convex functions in the following way. Suppose for simplicity that ρ is non-integer. Then take Tρ g = g  + ρ2 g and consider h1 (φ) := h2 (φ) :=

1 2ρ sin πρ 1 2ρ sin πρ



2π

c os ρ(φ − ψ − π)(Tρ g)+ (φ)dφ; 

0 2π

c os ρ(φ − ψ − π)(Tρ g)− (φ)dφ.

0

By Theorem 3.2.3.3 these functions are ρ trigonometrically convex and h1 −h2 = g. Hence (5.7.10.11), (5.7.10.12) holds.  Proof of Theorem 5.7.10.1. We consider a function f ∈ A(ρ(r)) with the limit set U := {v(z) = crρ h1 (φ) + (1 − c)rρ h2 (φ) : 0 ≤ c ≤ 1 } with h1 , h2 from the conditions of lemma, and we exploit Theorem 5.7.1.3.  Exercise 5.7.10.1 Do this in detail.

5.8. A generalization of the Valiron-Titchmarsh theorem

199

5.8 A generalization of the Valiron-Titchmarsh theorem 5.8.1 The point of departure on this topic is the following Theorem VT [Va, Ti] Let f ∈ A(ρ), ρ < 1 have its zeros on the negative ray. If the limit lim r−ρ log |f (r)| r→∞

exists, then the limit lim r−ρ n(r)

r→∞

exists. The latter means that f is a CRG-function. The general problem is the following. Let ρ be any non-integer number, f ∈ A(ρ(r)), and suppose all zeros of f lie on a finite system of rays KS1 := {z = reiφ : 0 < r < ∞, φ ∈ S1 }

(5.8.1.1)

S1 := {eiθj : j = 1, 2, . . . , m}.

(5.8.1.2)

where We write nf ∈ MS1 . Let nj be a zero distribution on the ray {arg z = θj } and all the limits lim r−ρ nj (r) := Δj

r→∞

(5.8.1.3)

exist. In such a case we write nf ∈ Mreg,S1 . Let KS be one more system of rays S = {eiψk : k = 1, 2, . . . , n}.

(5.8.1.4)

Some ψk can coincide with some θj . Suppose that f has regular growth on this system, i.e., (5.8.1.5) hf (φ) = hf (φ), eiφ ∈ S. In such a case we write f ∈ Areg,S . The problem is, what is the connection between S and S1 so that the implication (f ∈ Areg,S ) =⇒ (nf ∈ Mreg,S1 ) holds. This problem can be reformulated in another way. For nf ∈ Mreg,S1 if nf ∈ MS1 it is necessary and sufficient that f is a CRG-function, because existence of an angle density is equivalent to existence of all the limits. So the problem can be reformulated in the form: what is the connection between S and S1 , so that the implication (f ∈ Areg,S ) =⇒ (f is CRG-function) holds.

200

Chapter 5. Applications to Entire Functions

We write

G(t, γ, ρ) := Gp (et−iγ )e−ρt , p = [ρ]

(5.8.1.6)

where Gp is the Primary Kernel: Gp (z) = log |1 − z| + 

p 

z k /k.

k=1

Set

∞ ˆ γ, ρ) := G(s,

G(t, γ, ρ)e−ist dt.

−∞

This is the Fourier transformation of G(t, γ, ρ). It can be computed (see, e.g,[Oz, Lem. 3]); ˆ γ, ρ) = π cos(π + γ)(ρ + is) . G(s, (ρ + is) sin π(ρ + is) Consider the matrix ˆ S1 − S) := G(s, ˆ θj − ψk , ρ) . G(s,

(5.8.1.7)

We are going to prove (see [Az(1998)]) Theorem 5.8.1.1 The implication {f ∈ Areg,S } ∧ {nf ∈ MS1 } =⇒ {f is a CRG-function} holds iff ˆ S1 − S) = m, ∀s ∈ (−∞, ∞). rank G(s,

(5.8.1.8)

As a corollary we obtain the following ([De]) Theorem 5.8.1.2 (Delange) Suppose that S1 and S consist of one ray, i.e., S1 = {eiθ1 }, S = {eiψ1 }. The implication (5.8.1.5) holds iff θ1 − ψ1 = (1 − (2k + 1)/2ρ)π, k = 1, 2, . . . .

(5.8.1.9)

5.8.2 A Fourier transformation for distribution ν on the real axes is a distribution in the standard space S  (see [H¨o, vol. 1, Ch. 7, § 7.1]). For a locally bounded measure whose variation is “not very quickly” growing, it can be defined by ∞ (F ν)(s) := lim

→0 −∞

eits e−

t2 2

ν(dt)

where the right side is understood in the sense of distributions. For example, if ν(dt) := eis0 t dt, we have F ν(s) = δ(s − s0 ) where δ is the Dirac function.

5.8. A generalization of the Valiron-Titchmarsh theorem

201

Exercise 5.8.2.1 Check this. For distribution and a summable function one can define a convolution for which the property F (f ∗ ν)(s) = F f (s)F ν(s) holds. Proof of Theorem 5.8.1.1. Since f ∈ MS1 the limit set Frnf is concentrated on KS1 . So every v ∈ Fr[f ] can be represented in the form (see Theorem 3.1.5.1): v(z) =

∞ j=m 

Gp (z/reiθj )μj (dr)

(5.8.2.1)

j=1 0

where μj is concentrated on the ray {arg ζ = θj } and belongs to U [ρ]. After changing variables, r = eτ , |z| = et , we obtain from (5.8.2.1) 1

iφ

v (te ) =

∞ j=m  

G(t − τ, φ − θj , ρ)μ1j (dτ )

(5.8.2.1a)

j=1 −∞

where eρτ μ1j (dτ ) := μj (dr), v 1 (teiφ ) := v(|z|eiφ )e−ρ|z| .

(5.8.2.2)

The equality (5.8.2.1a) can be written as 1

iφ

v (te ) =

j=m 

[G(•, φ − θj , ρ) ∗ μ1j ](t)

(5.8.2.3)

j=1

where * stands for convolution. Then f ∈ Areg,S with nf ∈ MS1 , iff every pair v1 , v2 ∈ Fr[f ] satisfies the condition v1 (z) = v2 (z), z ∈ KS .

(5.8.2.4)

Denote by μ1,j , μ2,j the restriction of μv1 μv1 to the ray {arg z = θj }. Set νj := μ1,j − μ2,j Using (5.8.2.3) we can rewrite (5.8.2.4) in the form j=m 

[G(•, φk − θj , ρ) ∗ νj1 ](t) ≡ 0, k = 1, 2, . . . , n.

(5.8.2.5)

j=1

Applying Fourier transforms we obtain a system of linear equations: j=m  j=1

ˆ ψk − θj , ρ) · νˆj1 ](t) ≡ 0, k = 1, 2, . . . , n. [G(•,

(5.8.2.6)

202

Chapter 5. Applications to Entire Functions

ˆ S−S1 ) = m for every s ∈ R. The system (5.8.2.6) Suppose now that rank G(s, has only the trivial solution for every s. Thus νˆj1 (s) ≡ 0, for j = 1, 2, . . . , m. This implies νj1 (t) ≡ 0 for j = 1, 2, . . . , m and νj ≡ 0 for j = 1, 2, . . . , m. Thus μv1 = μv2 , i.e., (by (5.8.2.3)) Fr[f ] consists of one function v ∈ U [ρ]. Thus f is a CRG-function. ˆ S − S1 ) < m for some s0 . Conversely, suppose that rank G(s, Then there exists a nontrivial solution (b1 , . . . , bm ) that satisfies the corresponding system. We obtain that {ˆ νj1 bj δ(s − s0 ), j = 1, 2 . . . , m} is a solution of (5.8.2.6) for all s ∈ R and hence νj1 (dt) = bj eits0 dt, j = 1, 2, . . . , m. Since νj1 have bounded densities dνj1 /dt, we can find a constant C such that sup{|dνj1 /dt| : 0 < t < ∞, j = 1, 2 . . . , m} ≤ C. Set μ11,j (dt) = Cdt + νj1 (dt); μ12,j = Cdt. (5.8.2.7) Both of these are measures. Now we pass to m1,j , m2,j via (5.8.2.2). It is easy to check that m1,j , m2,j ∈ M(ρ). Exercise 5.8.2.2 Check this. Consider μ1 , μ2 ∈ M(ρ) which are defined uniquely by their restrictions μ1,j , μ2,j respectively on KS1 . Set   v1 (z) := Gp (z/ζ)μ1 (dξdη); v2 (z) := Gp (z/ζ)μ2 (dξdη); ζ = ξ + iη. C

C

It is easy to check that the equality v1 (z) = v2 (z), z ∈ KS

(5.8.2.8)

holds. Exercise 5.8.2.3 Check this. Since μ1 and μ2 are finite sums of trigonometrical functions, for v1 and v2 the condition (4.1.3.3) is satisfied. Thus by Theorem 4.3.6.1 there exists a function f ∈ A(ρ(r)) for which  C(cv1 + (1 − c)v2 ). Fr[f ] = 0≤c≤1

Since for v ∈ C(cv1 + (1 − c)v2 ) (5.8.2.8) also holds, the same holds for v ∈ Fr[f ] and this function is not a CRG-function. 

Chapter 6

Application to the Completeness of Exponential Systems in Convex Domains and the Multiplicator Problem The completeness of exponential systems in convex domains is intimately connected to the multiplicator problem. Considering a special form of exponent system is related to the study of special subharmonic functions that determine the periodic limit set, the so-called automorphic subharmonic functions. The next Sections 6.1, 6.2 are devoted to these problems.

6.1 The multiplicator problem 6.1.1 Let Φ ∈ A(ρ(r)) and let H(φ) be a ρ-trigonometrically convex function. A function g ∈ A(ρ(r)) is called an H-multiplicator of Φ if the indicator hgΦ of the product gΦ satisfies the inequality hgΦ (φ) ≤ H(φ), ∀φ. In some questions (see Section 6.3) we need to determine whether a given function Φ has a multiplicator. We shall study this problem in terms of the limit set of Φ. Define H(z) := rρ H(φ), z = reiφ . Let v ∈ U [ρ] (see (3.1.2.4)). Consider the function m(z, v, H) := H(z) − v(z). As will be proved in Corollary 6.1.9.3, the maximal subharmonic minorant of m(z, v, H) exists and is continuous. The maximal subharmonic minorant of m

204

Chapter 6. Application to the Completeness of Exponential Systems

(m.s.m.) belonging to U [ρ] will be denoted by GH v, while the domain of definition of the operator GH will be denoted by DH . Though m(0, •, •) = 0, the m.s.m. of m can differ from zero (as was remarked by A.E. Eremenko and M.L. Sodin), but if the m.s.m. of m equals zero at zero, then it belongs to U [ρ]. Exercise 6.1.1.0 Prove this. Exercise 6.1.1.1 Consider the function  |z| log |z|, if |z| ≤ 1; w(z) = |z| − 1, if |z| ≥ 1. It is subharmonic and belongs to U [1]. Show that the maximal subharmonic minorant of K|z| − w(z) is different from zero in 0 for every K > 0. Theorem 6.1.1.1 ([AG(1992)]) Φ ∈ A(ρ(r)) has an H-multiplicator iff Fr[Φ] ⊂ DH .

(6.1.1.1)

Proof of necessity. Let g be a multiplicator of Φ, i.e., hgΦ (φ) ≤ H(φ)

(6.1.1.2)

and let v ∈ Fr[Φ]. We can choose vgΦ ∈ Fr[gΦ] and vg ∈ Fr[g] such that vgΦ = v + vg (see Theorem 3.1.2.4, fru1)). Exercise 6.1.1.2 Prove this directly. By definition of indicator (3.2.1.1) and (6.1.1.2) we have vgΦ (z) ≤ H(z) or  vg (z) ≤ m(z, v, H). Since vg ∈ U [ρ], v ∈ DH . For proving sufficiency we need the following Theorem 6.1.1.2 The operator GH is 1. upper semicontinuous in the D -topology, 6.1.1.5, i.e., (vj → v) ∧ (GH vj → w) =⇒ (w ∈ U [ρ]) ∧ (w(z) ≤ GH (z), z ∈ C); 2. invariant: (GH v)[t] = GH v[t] ; (see (3.1.2.4a) for Pt ≡ tI); 3. concave: (∀v1 , v2 ∈ DH , c ∈ [0; 1]) =⇒ (vc := cv1 + (1 − c)v2 ∈ DH ) and GH (vc ) ≥ cGH (v1 ) + (1 − c)GH (v2 ).

6.1. The multiplicator problem

205

Proof. Let us prove 1). Suppose vj ∈ U [ρ] → v and GH vj → w. Then GH vj ≤ H(z) − vj (z), z ∈ C.

(6.1.1.3)

Applying (•) from (2.6.2.2) and Theorem 2.3.4.5, reg 3), we obtain w ≤ (H) (z) − (v) (z), z ∈ C w (0) ≥ 0. Passing to the limit as ↓ 0 we obtain by Theorem 2.6.2.3, ap2) w(z) ≤ H(z) − v(z) = m(z, H, v), z ∈ C. Since 0 ≤ w(0) ≤ m(0, H, v) = 0 we have w(0) = 0 and, hence, w ∈ U [ρ]. Thus v ∈ DH and w(z) ≤ GH v(z). Let us prove 2). Since H(z) is invariant with respect to (•)[t] , (Gv)[t] ≤ H(z) − v[t] . Hence, (Gv)[t] (z) ≤ (G(v[t] ))(z),

(6.1.1.4)

because G(v[t] ) is the maximal subharmonic minorant. We can replace v with v[1/t] and obtain (Gv[1/t] )[t] (z) ≤ Gv(z). Applying (•)[t] to the two sides of the inequality, we obtain Gv[1/t] (z) ≤ (Gv(z))[1/t] . Now we can replace 1/t with t and obtain the reverse inequality to (6.1.1.4), which, together with (6.1.1.4), proves 2). 3). Let v1 , v2 ∈ DH and c ∈ [0; 1]. One has Gvi (z) ≤ H(z) − vi (z), i = 1, 2, ∀z. Then [cGv1 + (1 − c)Gv2 ](z) ≤ H(z) − [cv1 + (1 − c)v2 ](z). Thus [cGv1 + (1 − c)Gv2 ](z) ≤ G[cv1 + (1 − c)v2 ](z).



Proof of sufficiency in Theorem 6.1.1.1. Assume that Fr[Φ] ⊂ DH and consider the set (6.1.1.5) U := {(v  , v  ) : v  ≤ Gv  , v  ∈ Fr[Φ]}. Then U is nonempty, because of (6.1.1.1), closed, because of Theorem 6.1.1.2, 1), and invariant, because of Theorem 6.1.1.2, 2). Every fiber U  = {v  : v  ≤ Gv  } is convex because of Theorem 6.1.1.2, 3). By Theorem 4.4.1.2 there exists u ∈ U (ρ(r)) such that for the curve u := (u , u ), Fr[u] = U .

(6.1.1.6)

By Theorem 5.3.1.4 (Approximation Theorem) the function u can be replaced with log |g|, where g ∈ A(ρ(r)), retaining the property (6.1.1.6).

206

Chapter 6. Application to the Completeness of Exponential Systems

Let us prove that g is an H-multiplicator of Φ. Indeed, set Π := gΦ. It is enough to prove that for every vΠ ∈ Fr[Π], vΠ (z) ≤ H(z).

(6.1.1.7)

Note that every vΠ has the form vΠ = vg + v, where (v, vg ) ∈ U . Thus, because of definitions (6.1.1.5) and (6.1.1.6), vΠ satisfies (6.1.1.7).  Let us note that the pair (v, GH v) ∈ U because of closeness of U . Hence the following assertion holds. Proposition 6.1.1.3 Every Φ ∈ A(ρ) that satisfies (6.1.1.1) has a multiplicator g ∈ A(ρ) such that (6.1.1.8) v + GH v ∈ Fr[gΦ]. Exercise 6.1.1.3 Check this in detail. Although v ∈ U [ρ] is in general an upper semicontinuous function, we need Theorem 6.1.1.4 The function GH v(z), v ∈ U [ρ], is a continuous function that is harmonic outside the set E = {z : GH v(z) = m(z, v, H)}. Proof. GG v(z) is continuous because of Corollary 6.1.9.3. If GH v(z0 ) < v(z0 ) and if GH v(z) is not harmonic in a neighborhood of z0 , we can make sweeping of masses from a small disc {|z − z0 | < } (see Theorem 2.7.2.1). The obtained subharmonic function will be greater than GH v(z), contradicting maximality.  6.1.2 Suppose that some H-multiplicator g = g(z, Φ, H) of the function Φ is found. We examine the function Π = gΦ. The structure of its limit set is described by the following statement: Proposition 6.1.2.1 Every vΠ ∈ Fr[gΦ] can be written as vΠ = v + w1 , where v ∈ Fr[Φ] and w1 ∈ U [ρ] with the condition w1 (z) ≤ GH (z), ∀z ∈ C,

(6.1.2.1)

and, conversely, for every v ∈ Fr[Φ] there exists a vg , vg (z) ≤ GH v(z), such that v + vg ∈ Fr[gΦ]. Exercise 6.1.2.1 Prove this the same way as in Exercise 6.1.1.2. An H-multiplicator G of the function Φ will be called ideally complementing if it satisfies the condition Fr[GΦ] = {vΠ = v + GH v : v ∈ Fr[Φ]}.

6.1. The multiplicator problem

207

If a multiplicator is ideally complementing, then equality is achieved in (6.1.2.1) for all v ∈ Fr[Φ]. This make the multiplicator optimal in another respect. Recall that an entire function f is of minimal type with respect to a proximate order ρ(r), ρ(r) → ρ if (see (2.8.1.6)) σf := lim sup log M (r, f )r−ρ(r) = 0. r→∞

Proposition 6.1.2.2 Let G = G(•, Φ, H) be an ideally complementing H-multiplicator of a function Φ. Then each H-multiplicator of the function Π = GΦ is of minimal type. This proposition is proved in Section 6.1.3. A function Φ is said to be ideally complementable if for each H the condition (6.1.1.1) implies that Φ has an ideally complementing multiplicator. For instance, if Φ is a function of completely regular growth (see Section 5.6) then it is ideally complementable. Exercise 6.1.2.2 Prove this. Theorem 6.1.2.3 Every function with periodic limit set is ideally complementable. This theorem is proved in Section 6.1.6. Let C ⊂ Rl be an l-dimensional connected compact and let {h(φ, c) : c ∈ C} be a set of ρ-t.c. functions that is continuous with respect to c ∈ C. For example, c ∈ [0, 1] and h(φ, c) = ch1 (φ) + (1 − c)h2 (φ). The set Uind := {v(reiφ s) = rρ h(φ, c) : c ∈ C}

(6.1.2.2)

is the limit set of an entire function. Exercise 6.1.2.3 Prove this using Theorem 4.3.6.1. Such a set is called a set of indicators. Entire functions with such limit sets can be also considered as a generalization of CRG-functions. Theorem 6.1.2.4 Every function Φ whose limit set is a set of indicators is ideally complementable. This theorem is proved in Section 6.1.7. The existence of an ideally complementing H-multiplicator depends, of course, both on Φ ∈ A(ρ(r)) (or, more precisely, on its limit set Fr[Φ]) and on H. Theorem 6.1.2.5 Let Φ and H be such that the condition (6.1.1.1) is satisfied. The function Φ has an ideally complementing H-multiplicator if and only if the operator GH is continuous on FrΦ. This theorem is proved in Section 6.1.6.

208

Chapter 6. Application to the Completeness of Exponential Systems

Now we formulate a sufficient condition for continuity of the operator GH . We shall say that the maximum principle for U [ρ] is valid in the domain G, (which is, generally speaking, unbounded), if the conditions w ∈ U [ρ], w(z) = 0 for z ∈ /G imply w(z) ≡ 0. Let us denote by Hw a region of harmonicity of w ∈ U [ρ], i.e., a region where the conditions “w is harmonic in G” and “G ⊃ Hw ” imply G = Hw . We remark that Hw is a connected component of the open set on which w is harmonic. Generally it is not unique. The image of U ∈ U [ρ] will be denoted by GH U , while its closure in the D -topology will be denoted by clos GH U . Theorem 6.1.2.6 Suppose for every w ∈ clos GH U and every Hw the maximum principle for U [ρ] holds. Then GH is continuous on U . This theorem is proved in Section 6.1.5. In Section 6.1.8 we will construct an example of Φ and H such that the operator GH is not continuous on Fr[Φ]. This is also an example of the function that has no ideally complementing multiplicator. 6.1.3 Proof of Proposition 6.1.2.2. Let g be an ideally complementing multiplicator of the function Π = GΦ. We write θ(z) := (gGΦ)(z).

(6.1.3.1)

Let vg ∈ Fr[g]. Let us choose tj → ∞ such that: (log |g|)tj → vg ; (log |Π|)tj → vΠ ∈ Fr[Π]; (log |θ|)tj → vθ ∈ Fr[θ]. It follows from (6.1.3.1) that vθ = vg + vΠ . Since g is a multiplicator of Π, we have vθ (z) = vg (z) + vΠ (z) ≤ H(z).

(6.1.3.2)

Since G is an ideally complementing multiplicator, vΠ = v + GH v. So for all z ∈ C (6.1.3.2) implies (vg + GH v)(z) ≤ (H − v)(z). Since GH v is the maximal subharmonic minorant, vg (z) ≤ 0 and hence vg (z) ≡ 0. Thus (see (3.2.1.1)) we have hg (φ) ≡ 0 and therefore σg = max hg (φ) = 0. 0≤φ≤2π



6.1.4 In order to prove Theorem 6.1.2.6 we need a number of auxiliary statements. Lemma 6.1.4.1 Let the maximum principle be valid in G for U [ρ] and for some continuous functions w1 , w ∈ U [ρ] satisfy: a) w is harmonic in G; b) w1 (z) = w(z) outside of G.

6.1. The multiplicator problem

209

Then w1 (z) ≤ w(z), z ∈ G. Proof. We set

 w0 (z) :=

(w1 − w)+ (z), 0,

(6.1.4.1)

z ∈ G, z∈ / G.

This function is continuous in C and, evidently, subharmonic both in G and in C \ G. Since w0 (z) ≥ 0, the inequality for the mean values 1 0 = w0 (z) ≤ 2π

2π w0 (z + reiφ )dφ, z ∈ ∂G, 0

implies the subharmonicity on ∂G. Since G satisfies the maximum principle for U [ρ], we have w0 ≡ 0, which is equivalent to (6.1.4.1).  Now we shall dwell on some properties of maximal subharmonic minorants and, in particular, of w = GH v. Let Ev := {z ∈ C : GH v(z) = m(z, v, H)}.

(6.1.4.2)

We remark that m(z, v, H) is a δ-subharmonic function in C whose charge will be denoted by ν(•, v), its positive and negative parts will be denoted by ν + and ν − . Let us denote by μH the measure of H(z). It is decomposed into the product of measures (see Section 3.2 and Proposition 5.6.3.1) μH = ΔH ⊗ ρrρ−1 dr,

(6.1.4.3)

where ΔH is the measure on the unit circle and ρrρ−1 dr is the measure on the ray. It is obvious that (6.1.4.4) ν + (•, v) ≤ μH (•). We shall denote the mass distribution of w ∈ U [ρ] by μw . The modulus of continuity of w (if w is continuous) will be denoted by ωw (z, h), z ∈ C, h > 0. The following lemma lists various properties of w ∈ GH U, U ⊂ U [ρ] which will be useful in the sequel: Lemma 6.1.4.2 Let w ∈ GH U . Then 1. w ∈ U [ρ, σ] where σ = 4 · 2ρ [max{H(eiφ ) : φ ∈ [0, 2π]} + 2σ1 ], σ1 = max{v(z)|z|−ρ : z ∈ C, v ∈ U };

210

Chapter 6. Application to the Completeness of Exponential Systems

2. the charge restriction ν(•, v)|Ev to Ev is nonnegative, i.e., ν(•, v)|Ev = ν + (•, v)|Ev ; 3. outside Ev the function w is harmonic, i.e., μw |C\Ev = 0; 4. the measure μw is bounded from above by ν + (•, v), i.e., μw ≤ ν + (•, v); 5. GH U is equicontinuous on each compact set, i.e., √ ωw (z, h) ≤ C(R, σ, ρ) h log(1/h), |z| ≤ R, where C(R, σ, ρ) is independent of w ∈ GH U . Proof. Let us prove property 1. We have 1 T (r, w) := 2π

2π w+ (reiφ )dφ 0

1 ρ ≤ r 2π

2π

2π +

iφ

H (e )dφ + 0

2π +

iφ

v (re )dφ + 0

 v − (reiφ )dφ .

0

Since v(0) = 0, we have 2π

−

2π

v (re )dφ ≤ 0

iφ

v + (reiφ )dφ. 0

Therefore T (r, w) ≤ [max{H(eiφ ) : φ ∈ [0, 2φ]} + 2σ1 ]rρ . It is known (see Theorem 2.8.2.3, (2.8.2.5)) that M (r) ≤ 4T (2r). So we conclude that w(z) ≤ 4 · 2ρ [max{H(eiφ ; φ ∈ [0, 2π]} + 2σ1 ]|z|ρ = σ|z|ρ . Let us prove property 2. To this end we shall use the following theorem (Grishin’s Lemma) [Gr]. Theorem A.F.G Let g be a nonnegative δ-subharmonic function, and let νg be its charge. Then the restriction νg |E to the set E = {z : g(z) = 0} is a measure. Applying this theorem to the function g := m(z, v, H) − GH v(z), we get ν(•, v)|Ev ≥ μw |Ev , hence, we obtain property 2.

(6.1.4.5)

6.1. The multiplicator problem

211

Let us prove property 3. Since w and v are upper semicontinuous, and H is continuous (see Theorem 3.2.5.5), the set {z : (w + v)(z) − H(z) < 0} is open. Let us take a neighborhood of an arbitrary point of this set and replace the function w within it with the Poisson integral constructed using this function, i.e., let us sweep out the mass from this neighborhood. The subharmonic function obtained would be strictly greater than the initial one, if the latter were not harmonic. This means that the initial w was not the maximal minorant. We have arrived at a contradiction, which proves property 3. Property 4 immediately follows from property 3 and (6.1.4.5). In order to prove property 5 we shall need an auxiliary statement which will be stated as a number of lemmas. Let P (z, φ, R) :=

1 R2 − |z|2 2π |z − Reiφ |

be the Poisson kernel in the disc KR = {z : |z| < R}. Below, C  s with indices will denote constants. Lemma 6.1.4.3 In the disc KR/2 , we have | gradz P (z, φ, R)| ≤ C1 (R), where C1 (R) depends only on R. Exercise 6.1.4.1 Prove this. We shall introduce the notation for the Green function for the Laplace operator in the disc KR :   2  R − ζz  .  G(z, ζ, R) := log  R(z − ζ)  The disc {ζ : |ζ − z| < t} will be denoted by Kz,t . Lemma 6.1.4.4 Let z ∈ KR/2 \ Kζ,√h . Then for a small h, √ | gradz G(z, ζ, r)| ≤ C2 (R)/ h. Exercise 6.1.4.2 Prove this. Let us write μ(z, t) := μ(Kz,t ). Lemma 6.1.4.5 For z ∈ KR/2 , 0 < t < R/10, we have μH (z, t) ≤ C3 (σ, R)t.

212

Chapter 6. Application to the Completeness of Exponential Systems

Proof. Applying Theorem 2.6.5.1 (Jensen-Privalov) to the function H(z), we obtain MH = max{H(eiφ : φ ∈ [0; 2π]} = ΔH (T)/ρ where T is the unit circle. Now |z|+t 

rρ−1 dr ≤ ρ2 MH Rρ−1 t ≤ σC(ρ)Rρ−1 t

μ(z, t) ≤ ΔH (T) |z|−t

where C(ρ) is a constant depending only on ρ. This proves the lemma.



Lemma 6.1.4.6 Let h < 1 and suppose that a monotonic function μ(t) satisfies the condition μ(t) < ct. (6.1.4.6) Then

√

h √ log(1/t)μ(dt) ≤ (3/2)c h log h. 0

Exercise 6.1.4.3 Prove this by integrating by parts and using (6.1.4.6). Lemma 6.1.4.7 Let z ∈ KR/2 and ζ ∈ KR . Then | log |(R2 − ζz/R|| ≤ C4 (R). Exercise 6.1.4.4 Prove this. Now we pass to the proof of assertion 5 from Lemma 6.1.4.2. According to the F. Riesz theorem (Theorem 2.6.4.3) we represent w in the circle as  w(z) = H(z, w) − G(z, ζ, R)μw (dξdη), ζ = ξ + iη, (6.1.4.7) KR

where 1 H(z, w) = 2π

2π P (z, φ, R)w(Reiφ )dφ. 0

It follows from Lemma 6.1.4.3 and 1 of Lemma 6.1.4.2 that 1 | gradz H(z, w)| ≤ C1 (R) 2π

2π |w|(Reiφ )dφ ≤ C1 (R)2σRρ . 0

(6.1.4.8)

6.1. The multiplicator problem

213

We split the integral in (6.1.4.7) into three terms:  ψ1 (z, h) := G(z, ζ, R)μw (dξdη), Kr \Kz



√ 0, h

log |(R2 − zζ)/R|μ(dξdη),

ψ2 (z, h) = Kz

√ 0, h



log |ζ − z|μ(dξdη),

ψ3 (z, h) = Kz

√ 0, h

where z0 is an arbitrary fixed point in KR/2 . Combining property 4 and inequality (6.1.4.4) we have μw (E) ≤ μH (E), ∀E ⊂ KR .

(6.1.4.9)

For all z ∈ Kz0 ,√h/2 Lemma 6.1.4.4 yields

√ | grad ψ1 (z, h)| ≤ C2 (r)σRρ / h.

(6.1.4.10)

Combining Lemmas 6.1.4.5 and 6.1.4.7 with inequality (6.1.4.9), we get √ |ψ2 (z, h)| ≤ C4 (R)C3 (σ, R) h. (6.1.4.11) Further, from Lemmas 6.1.4.5 and 6.1.4.6, taking into account the fact that log |ζ − z| < 0, we obtain √ |ψ3 (z, h)| ≤ (3/2)C3 (σ, R) h log h. (6.1.4.12) Now consider the difference Δw := w(z0 + Δz) − w(z0 ), |Δz| < h <

√ h/2.

It can be represented as Δw = Δψ1 + Δψ2 + Δψ3 + ΔH(z, w). Choosing h small enough, one may assume that z0 + Δz ∈ cording to (6.1.4.11),

(6.1.4.13) K√h/2,z0 .

√ |Δψ2 (z0 , h)| ≤ |ψ2 (z0 , h)| + |ψ2 (z0 + h, h)| ≤ C6 (σ, R) h.

Thus, ac(6.1.4.14)

Likewise (6.1.4.12) yields

√ |Δψ3 (z0 , h)| ≤ |ψ3 (z0 , h)| + |ψ3 (z0 + h, h)| ≤ C7 (σ, R) h log h.

Finally, from (6.1.4.10) and (6.1.4.8), respectively, we obtain √ |Δψ1 | ≤ C3 (σ, R) h, |ΔH(z0 , w)| ≤ C8 (σ, R)h.

(6.1.4.15)

(6.1.4.16)

Substituting (6.1.4.14)–(6.1.4.16) into (6.1.4.13), we obtain relation 5 of Lemma 6.1.4.2.  Thus we have completed the proof of Lemma 6.1.4.2.

214

Chapter 6. Application to the Completeness of Exponential Systems

6.1.5 In this item we are going to prove Theorem 6.1.2.6. However, before that, we prove D

D

Lemma 6.1.5.1 Let wn = GH vn , vn → v and wn → w∞ . Set E∞ := {z : w∞ (z) = H(z) − v(z)}. Then w∞ is harmonic in C \ E∞ . Let us note that w∞ , in general, is not the maximal subharmonic minorant because the operator GH can be only upper semicontinuous, as will be demonstrated by example in Section 6.1.8. However, w∞ is a minorant of H − v because of Theorem 6.1.1.2, 1. / E∞ . Then there exists a δ > 0 such that Proof. Let z0 ∈ w∞ (z0 ) + v(z0 ) ≤ H(z0 ) − 2δ. Since the function b(z) := w∞ + v(z) − H(z) is upper semicontinuous, there exists an = (δ) such that b(z) < −δ for all z ∈ {|z − z0 | < 2 }. D

Let (•) be a smoothing operator from (2.6.2.3). If wn → w then (wn ) → w uniformly on every compact set (Theorem 2.3.4.5, reg3) and for every subharmonic function v the sequence v (z) ↓ v(z),when ↓ 0 (Theorem 2.6.2.3, ap2). Then (b) (z) < −δ, for |z − z0 | < or (w∞ ) (z) + (v) (z) ≤ (H) (z) − δ. The function H is continuous, hence uniformly continuous on the circle {z : |z − z0 | ≤

}. Thus we can replace (H) in the last inequality with H and δ with δ/2. So, we have (6.1.5.1) (w∞ ) (z) + v (z) ≤ H(z) − δ/2, |z − z0 | < . Since (•) is monotonic on subharmonic functions, we can replace in (6.1.5.1) with any 1 < . So we obtain (w∞ )1 (z) + v1 (z) ≤ H(z) − δ/2, |z − z0 | < .

(6.1.5.2)

Since (wn )1 → (w∞ )1 uniformly in the disc |z −z0 | ≤ we can replace in (6.1.5.2) w∞ with wn and respectively v with vn , changing δ/2 with δ/4. After that we can pass to the limit as 1 ↓ 0 for every sufficiently large n. So we obtain wn (z) + vn (z) ≤ H(z) − δ/4, |z − z0 | < . It means that the disc {|z − z0 | < } ⊂ C \ Evn . Because of Lemma 6.1.4.2, 3, wn is harmonic in this disc for all large n. Thus w∞ is also harmonic, as the D -limit of wn .  D

Proof of Theorem 6.1.2.6. Let vn → v. Then the set wn = GH vn is equicontinuous by Lemma 6.1.4.2, 5, and we can choose from it a subsequence uniformly converging to a continuous function w∞ . Let w = GH v, E = Ew , E∞ being defined in Lemma 6.1.5.1.

6.1. The multiplicator problem

215

Since (w∞ + v)(z) ≤ H(z), (w + v)(z) ≤ H(z), ∀z ∈ C and v is upper semicontinuous, whereas w and H are continuous, the sets E and E∞ are closed. Since w∞ (z) ≤ H(z) − v(z) we have w∞ (z) ≤ w(z), ∀z ∈ C,

(6.1.5.3)

and therefore E∞ ⊂ E. The function w is subharmonic in C \ E∞ , and w∞ is harmonic in C \ E∞ by Lemma 6.1.5.1. They take the same values on E∞ . As the maximum principle holds in Hw∞ by assumption we have, according to Lemma 6.1.4.1 the inequality w(z) ≤ w∞ (z), ∀z ∈ C.

(6.1.5.4)

The inequalities (6.1.5.4) and (6.1.5.3) imply that w(z) = w∞ (z), i.e., GH is continuous.  6.1.6 Proof of Theorem 6.1.2.5. Sufficiency. We exploit the following criterion for existence of a limit set that follows from Theorems 4.2.1.1, 4.2.1.2, 4.3.1.2 and Corollary 5.3.1.5: Proposition 6.1.6.1 In order that U ⊂ U [ρ] be a limit set of an entire function f ∈ A(ρ(r)) it is necessary and sufficient that there exists a piecewise continuous, ω-dense in U asymptotically dynamical pseudo-trajectory (a.d.p.t) v(•|t). Exercise 6.1.6.1 Check this. Let vΦ (•|t) be an a.d.p.t. corresponding to FrΦ. Consider the pseudo-trajectory vg (•|t) := GH vΦ (•|t). It exists because of (6.1.1.1). Prove that this pseudotrajectory is asymptotically dynamical, i.e., (4.3.1.1) is fulfilled. Recall that Tτ • = (•)[eτ ] . Using the property of invariance of GH (Theorem 6.1.1.2, 2) we have Tτ vg (•|et ) − vg (•|et+τ ) = GH [Tτ vΦ (•|et ) − vΦ (•|et+τ )]. Thus (4.3.1.1) is fulfilled because of continuity of GH . Also the condition of ωdenseness (4.3.1.4) is fulfilled and {w ∈ U [ρ] : (∃tj → ∞) w = D − lim vg (•|etj )} = GH (FrΦ). The corresponding entire function g ∈ A(ρ(r)) with the limit set Ug = GH (FrΦ) is an ideally complementing multiplicator, because Fr[gΦ] = {v + GH v : v ∈ Fr[Φ]}.

216

Chapter 6. Application to the Completeness of Exponential Systems

Exercise 6.1.6.2 Check this. Necessity. Let G be an ideally complementing multiplicator of Φ. Let us show that GH is continuous on Fr[Φ]. Assume this is not the case, i.e., there exists a sequence vj → v such that GH vj → W and W = GH v. Since the limit set Fr[GΦ] is closed, we have vj + GH vj → v + GH v, vj ∈ Fr[Φ]. On the other hand,  vj + GH vj → v + W . Thus, W = GH v, which is a contradiction. Proof of Theorem 6.1.2.3. Let Fr[Φ] be a periodic limit set, that is Fr[Φ] = C(v) = {v[t] : 1 ≤ t ≤ eP }, where v ∈ U [ρ]. We shall show that GH is continuous on U [ρ]. By Theorem 6.1.1.2, 2) the equality (GH v)[t] = GH v[t] holds. Since the operation (•)[t] is continuous for  all t, GH is continuous on C(v). 6.1.7 Now we are going to prove Theorem 6.1.2.4. However, we need some preparation. Let h(φ), φ ∈ [0, 2π) be a 2π-periodic ρ-t.c.function, satisfying the condition max h(φ) = σ. φ∈[0,2π]

We denote this class as T C[ρ, σ] and write T C[ρ] :=



T C[ρ, σ].

σ>0

The class of functions w = h1 − h2 where h1 , h2 ∈ T C[ρ, σ] will be denoted as δT C[ρ, σ] and we will also write δT C[ρ] :=



δT C[ρ, σ].

σ>0

From properties of a ρ-t.c.function (see Sections 3.2.3–3.2.5) we can obtain the following properties of δ − ρ-t.c.functions: Proposition 6.1.7.1 For w ∈ δT C[ρ] the following holds: 1. w (φ − 0) and w (φ + 0) exist at each point and are bounded in [0; 2π]; 2. w (φ − 0) = w (φ + 0) for all φ ∈ [0; 2π], except, perhaps, a countable set; 3. the charge Δw generated by the function 

Δw := w (φ) + ρ

 2

φ

w(θ)dθ

6.1. The multiplicator problem

217

has bounded variation |Δw |; the variation |Δw |(α, β) of the charge on the interval (α; β) and the variation of the charge generated by derivative |w |(α; β) on the same interval satisfy the relation |Δw |(α, β) ≥ |w |(α; β) + ρ2 (β − α). 4. For all w ∈ δT C[ρ, σ], max(|w (φ − 0)|, |w (φ + 0)|) ≤ C(ρ, σ), φ ∈ [0; 2π]; D

5. if rρ wn → rρ w and wn ∈ δT C[ρ, σ], then wn → w uniformly on [0; 2π]. Exercise 6.1.7.1 Prove this using properties of ρ-t.c.functions. We also need a technical Lemma 6.1.7.2 Let Mn (φ) be a sequence of functions that satisfy the conditions: 1. Mn ≥ 0; Mn (0) = 0; 2. Mn converges uniformly to M∞ (φ) ≥ A sin ρφ, A > 0; 3. Mn (φ − 0), Mn (φ + 0) exist at every point, and they coincide almost everywhere; 4. there exists a sequence φn ↓ 0 such that for each arbitrarily small > 0 and arbitrarily large n0 ∈ N there exists n > n0 for which the inequality Mn (φn ) < φn holds. Then there exists a sequence (ζn , ηn ) of disjoint intervals and a subsequence Mkn such that Mk n (ζn ) − Mk n (ηn ) ≥ Aρ/2. (6.1.7.1) Proof. Set 0 = 1/2, η0 = π/4 and choose the required sequence recurrently. Let n , ηn , ζn be already chosen. Set n+1 = n /2, find φn+1 < ηn and choose k0 = k0 (n) so that for k > k0 , Mk (φn+1 ) − Aρφn+1 > − n+1 φn+1 . This is possible because of condition 2 and sin ρφ ∼ ρφ, φ → 0. So we have Mk (φn+1 ) > Aρ − n+1 . φn+1

(6.1.7.2)

Now, choose ψn+1 < φn+1 and kn+1 > k0 so that Mkn+1 (ψn+1 ) < n+1 ψn+1 .

(6.1.7.3)

This is possible by condition 4. Thus for small n+1 from (6.1.7.2) and (6.1.7.3) we obtain Mkn+1 (φn+1 ) − Mkn+1 (ψn+1 ) > (2/3)Aρ. (6.1.7.4) φn+1 − ψn+1

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Chapter 6. Application to the Completeness of Exponential Systems

On the other hand

Mkn+1 (ψn+1 ) − Mkn+1 (0) < n+1 . ψn+1 − 0

(6.1.7.5)

On the interval (ψn+1 , φn+1 ) there is a point ηn+1 where the derivative exists and the inequality Mk n+1 (ηn+1 ) ≥

Mkn+1 (φn+1 ) − Mkn+1 (ψn+1 ) φn+1 − ψn+1

(6.1.7.6)

is valid. Also there is a point ζn+1 ∈ (0, ψn+1 ) where the derivative exists and the inequality Mkn+1 (ψn+1 ) − Mkn+1 (0) (6.1.7.7) Mk n+1 (ζn+1 ) ≤ ψn+1 − 0 is valid. From the inequalities (6.1.7.4)–(6.1.7.7) we obtain (6.1.7.1).



Proof of Theorem 6.1.2.4. Denote by GˆH h the maximal ρ-t.c.minorant of H(eiφ ) − h(φ). It follows from Theorem 6.1.1.2, 2 that GH (rρ h(φ))(reiφ ) = rρ GˆH h(φ). Exercise 6.1.7.2 Prove this. So taking in consideration Proposition 6.1.7.1, 5, one must prove Proposition 6.1.7.3 The operator GˆH is continuous on the set ˆind := {h(φ, c) : c ∈ C} U in the uniform topology. ˆind . Set w Proof. Let hn → h, hn , h ∈ U ˆn = GˆH hn , w ˆ = GˆH h, w ˆ∞ = limn→∞ w ˆn . ˆ ∞ = H −h−w ˆ = H − h − w. ˆ n = H − hn − w ˆn , M ˆ∞, M ˆ Let (αn ; βn ) We set also M ˆ n (φ) > 0. We shall show that βn − αn ≤ π/ρ. be a maximum interval where M Indeed, for a fixed n let us consider the function ˆ n := w W ˆn + n L(φ − (αn + βn )/2) where

 cos |φ|, L(φ) = 0,

φ ∈ (−π/2ρ; π/2ρ); φ ∈ [−π; π] \ (−π/2ρ; π/2ρ),

ˆ n is also a ρ-t.c.minorant of and n is small enough. If βn − αn > π/ρ, then W ˆn is not maximal. H − hn , i.e., w

6.1. The multiplicator problem

219

If βn − αn = π/ρ, then, to ensure that w ˆn is a maximal minorant, at least one of the conditions ˆ n (φ) ˆ n (φ) M M =0 = 0, lim inf φ→αn +0 φ − αn φ→βn −0 βn − φ lim inf

(6.1.7.8)

must be satisfied. ˆ n (φ) for Let us choose (and preserve the previous notation) a subsequence M which αn → α, and βn → β. If β − α < π/ρ, then the maximum principle for ρ-t.c.functions is valid. ˆ for all φ, Repeating arguments of proof of Theorem 6.1.2.6, we obtain w ˆ∞ = w which proves Proposition 6.1.7.3 for the case considered. Exercise 6.1.7.3 Repeat them. Consider the case when β − α = π/ρ. Set q(φ) = (w − w∞ )(φ). The function q is ρ-trigonometric on the interval (α; β) since w ˆ and w ˆ∞ are ρ-trigonometric, i.e., have the form A sin ρφ + B cos ρφ. Exercise 6.1.7.4 Explain this. Besides, we have q ≥ 0 and q(α) = q(β) = 0. It is easy to see that q has the form q(φ) = A sin ρ(φ − α), A > 0.

(6.1.7.9)

Exercise 6.1.7.5 Prove this. ˆ (φ)+(wˆ − w ˆ (φ) ≥ 0, we have M ˆ ∞ (φ) = M ˆ∞ )(φ) ≥ (wˆ − w ˆ∞ )(φ), ∀φ, Since M whence ˆ ∞ (φ) ≥ A sin ρ(φ − α), A > 0. M (6.1.7.10) Since βn − αn ≤ π/ρ, the segment [α, β] contains the infinite sequence αn or βn . Let us single out a subsequence, let it be, for example, αn → α + 0, αn ∈ [α; β]. ˆ n (φ − a). From the definition of Mn and Consider the sequence Mn (φ) = M from relation (6.1.7.10) it follows that conditions 1 and 2 of Lemma 6.1.7.2 are fulfilled. Condition 3 is fulfilled because of property 1 of Lemma 6.1.7.2. Further, if αn ≡ α, then condition 4 of Lemma 6.1.7.2 is trivially true,since Mn (αn − α) = 0; otherwise, if αn ≡ α, condition 4 follows from (6.1.7.8). Applying Lemma 6.1.7.2, we obtain the union of intervals satisfying (6.1.7.7). ˆ n − hn − w The equality H(φ) = M ˆn yields the following inequality for the measure ΔH :ΔH ((ηn ; ζn )) ≥ Aρ/2. Summing this inequality and taking into account the fact that the intervals do not intersect, we obtain ΔH (∪n (ηn , ζn )) = ∞, which is impossible. So, Proposition 6.1.7.3 is proved.  Hence, Theorem 6.1.2.4 is proved.



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Chapter 6. Application to the Completeness of Exponential Systems

6.1.8 In this item we show an example of H and an entire function without an ideally complementing H-multiplicator. According to Theorem 6.1.2.5, to construct such an example it is sufficient to construct a limit set on which GH is not continuous. We set  cos |η|, η ∈ (−π/2ρ; π/2ρ); L(η) = 0, η ∈ [−π; π] \ (−π/2ρ; π/2ρ). Let us define X ∈ C ∞ so that X(ξ) = 1 for ξ < 0 and X = 0 for ξ > α. We set κ := (1/ρ2 )

max [2ρX  + X  ](ξ), H0 (η) := L(η) + κ.

(6.1.8.1)

(−∞;+∞)

We also set v(ζ, c) := [H0 − X(ξ − c)L(η)]eρξ , ζ = ξ + iη, where H0 and L have been periodically extended from the interval [−π; π] to (−∞, +∞). As H(z) we take H(z) := H0 (φ)rρ . Lemma 6.1.8.1 We have v(log z, c) ∈ U [ρ, σ], σ = 1 + κ, GH v(•, c) ≡ 0,

(6.1.8.2) (6.1.8.3)

lim v(log z, c) = κrρ

(6.1.8.4)

c→∞

uniformly with respect to z ∈ K  C, and GH (κrρ ) = L(φ)rρ .

(6.1.8.5)

Proof. For the Laplace operators in ζ and z it is true that Δζ = Δz /|ζ|2 . Let us check that v(ζ, c) is subharmonic in ζ. We have Δζ v(ζ, c) = {[1−X(ξ −c)](L +ρ2 L)(η)+[ρ2 κ−L(ξ)[X  (x−c)+2ρX (x−c)]}eρξ . Exercise 6.1.8.1 Check this computation. Since X(ξ) ≤ 1 and L(η) is ρ-t.c., [1 − X(ξ − c)](L + ρ2 L)(η) ≥ 0. Since L(ξ) ≤ 1 and [X  (x − c) + 2ρX  (x − c)] ≤ κρ2 we have [ρ2 κ − L(ξ)[X  (x − c) + 2ρX  (x − c)] ≥ 0. Thus v(log z, c) is subharmonic.

6.1. The multiplicator problem

221

Exercise 6.1.8.2 Prove that v(log z, c) ∈ U [ρ, σ] for σ = 1 + κ. Let us prove (6.1.8.3). We have H(z) − v(log z, c) = X(log r − c)L(φ)rρ . Since X = 0 for r > ec+α , the maximal subharmonic minorant of H − v is zero by the maximum principle. Relation (6.1.8.4) is obvious, since X(log r − c) converges to 1 uniformly on every disc {|z| ≤ R}. Relation 6.1.8.5 follows from the equality H(z) − κrρ = L(φ)rρ , since L(φ)rρ ∈ U [ρ].  Now we pass to the construction of the example. Examine the set U1 := clos{v(log z, c) : c ∈ [0; ∞)}. It contains the function D − lim v(log z, c) = κrρ . c→∞

Let us consider the minimal convex (•)[t] -invariant set U containing U1 . The set is contained in U [ρ, 1 + κ]. It is a limit set for a certain entire function Φ. Let us show that GH is not continuous on Fr[Φ]. We take an arbitrary sequence cj → ∞ and set vj (z) := v(log z, cj ) ∈ U . Now D − limj→∞ vj = κrρ by (6.1.8.4) and GH vj (z) = 0, so D − limj→∞ GH vj = 0 but GH (lim vj ) = GH (κrρ ) = L(φ)rρ ≡ 0 which shows the lack of continuity. By virtue of Theorem 6.1.2.5, Φ is not ideally complementable. 6.1.9 Here we prove existence and continuity of the maximal subharmonic minorant for some classes of functions m(z) . Theorem 6.1.9.1 Let m(z) be a continuous function such that the set of subharmonic minorants is nonempty. Then the maximal subharmonic minorant of m exists and is continuous. Proof. The set of subharmonic minorants is nonempty and partially ordered. Indeed, for every subset {uα , α ∈ A} of subharmonic minorants there exists uA = (sup{uα : α ∈ A})∗ which is subharmonic and is a minorant of m, because m is continuous. Exercise 6.1.9.1 Explain this in detail.

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Chapter 6. Application to the Completeness of Exponential Systems

Thus there exists a uniquely maximal element m.s.m.(z, m), which is a subharmonic minorant of m. Let us prove that it is continuous at every point z0 . Since m.s.m.(z, m) is upper semicontinuous, m.s.m.(z, m) > m.s.m.(z0 , m) − for |z − z0 | < δ for arbitrarily small and corresponding δ = δ( ). So we need to prove the inequality m.s.m.(z, m) < m.s.m.(z0 , m) + for arbitrarily small and corresponding δ = δ( ). Perform sweeping m.s.m.(z, m) from the disc |z − z0 | < δ such that the result u(z, δ) satisfies the inequality m.s.m.(z, m) < u(z, δ) < m.s.m.(z, m) + < m(z) + . Thus u(z, δ) − < m(z). Hence m.s.m.(z, m) > u(z, δ) − for all z. Since u(z, δ) is continuous, u(z, δ) > u(z0 , δ) − in the disc {|z − z0 | < δ1 }. So m.s.m.(z, m) > u(z0 , δ) − > m.s.m.(z0 , m) − .  Theorem 6.1.9.2 Let m = m1 − m2 , where m1 , m2 are subharmonic functions. Then the maximal subharmonic minorant of m exists. If m1 is continuous, then the maximal subharmonic minorant is continuous. Proof. Set M (z, m) := M (z, m1 ) − M (z, m2 ), where M (z, mi ), i = 1, 2 is defined by (2.6.1.1). Since M (z, m) is continuous (see Theorem 2.6.2.3 (Smooth approximation)), there exists m.s.m.(z, M (z, m)). We have u(z, m) := lim sup m.s.m.(z, M (•, m)) ≤ lim M (z, m) = m1 − m2 (z) = m(z). →0

→0

Now we prove that the upper semicontinuous regularization u∗ (z, m) also satisfies the inequality u∗ (z, m) ≤ m(z). Indeed, m2 + u(z, m) ≤ m1 (z). Hence, M (z, m2 ) + M (z, u(•, m)) ≤ M (z, m1 ). Passing to the limit we obtain three subharmonic functions and inequality m2 (z) + u∗ (z, m) ≤ m1 (z). We prove that u∗ (z, m) is the m.s.m.(z, m). If not, there would exist a subharmonic function u1 which exceeds u∗ (z, m) on a set of positive measure (otherwise they coincide); thus we would have for some z and , u∗ (z, m) < M (z, u1 ) ≤ m.s.m.(z, M (z, m)). This contradicts the definition of u∗ (z, m).

6.2. A generalization of ρ-trigonometric convexity

223

Now suppose that m1 is continuous at a point z0 . From Theorem 2.6.5.1 (Jensen-Privalov) we obtain that it is equivalent to   μm1 ({z : |z − z0 | < t}) dt = o(1), → 0. t 0 Similarly to the proof of Theorem 6.1.4.2, 5, we obtain μm.s.m.(z,m) ≤ μm1 . 

Hence m.s.m.(z, m) is also continuous. Exercise 6.1.9.2 Prove continuity in detail.

Corollary 6.1.9.3 For m = m(z, v, H) the function GH v(z) := m.s.m.(z, m) exists and is continuous. Exercise 6.1.9.3 Prove Corollary 6.1.9.3.

6.2 A generalization of ρ-trigonometric convexity 6.2.1 One of the important and useful kinds of limit sets is periodic limit sets. They are determined by one subharmonic function v ∈ U [ρ] that satisfies the condition (6.2.1.1) v(T z) = T ρ v(z), z ∈ C. Such a function is called automorphic. They generate the class of so-called Lρ subfunctions, that is a generalization of ρ-trigonometrically convex functions. In this part we are going to review properties of such functions from different points of view that will be useful for applications (see [ADP]). In connection with property (6.2.1.1) it is natural to consider so-called T homogeneous domains in C, i.e., such domains G that satisfy the condition {T z : z ∈ G} = G or shortly T G = G. As we can see they are invariant with respect to dilation by T . For example, every component of an open set of harmonicity of an automorphic function is a T -homogeneous domain. Let v satisfy (6.2.1.1). Then the function q(z) := v(ez )e−ρx

(6.2.1.2)

is a 2π periodic function in y and P -periodic in x, where P = log T . The function q can be considered as a function on a torus T2P , obtained by identifying the opposite sides of the rectangle Π = (0, T ) × (−π, π). The homology group of T2P is nontrivial, and generated by the cycles γx , γy , where γx = T2P ∩ {y = 0}, γy = T2P ∩ {x = 0}.

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Chapter 6. Application to the Completeness of Exponential Systems

Let π be the covering map of C onto T2P , then φ = π ◦ log is a well-defined covering map of C \ {0} onto T2P , where the group of deck transformations is given by the dilations by T m for m ∈ Z. So if G is a given T -homogeneous domain, then D = π ◦ log G = φ(G)

(6.2.1.3)

is a domain in T2P . On the other hand, not every domain in T2P has a T -homogeneous domain as its preimage under φ. The preimage φ−1 (D) under φ is a possibly disconnected set which is invariant under dilations by T m for m ∈ Z. An intrinsic description is given by the next proposition. Proposition 6.2.1.1 Let γ be a closed curve in a domain D ⊂ T2P that is homologous in T2P to a cycle γ = nx γx + ny γy , nx , ny ∈ Z. 1. If nx = 0 for every such γ in D, then φ−1 (D) = ∪∞ j=−∞ Gj , where Gj = T j G0 , G0 is an arbitrary connected component of φ−1 (D), and Gj ∩ Gl = ∅ for j = l. 2. If there exists a curve γ as above with nx = 0, then φ−1 (D) = ∪k−1 q=0 Gq , where k = min |nx | with the minimum taken over all such curves γ; G0 is an arbitrary component of φ−1 (D); Gj , j = 0, 1, . . . , k − 1, are disjoint T k -homogeneous domains, and for every m ∈ Z, T m G0 = Gq , provided m = lk + q, for some q ∈ Z, 0 ≤ q ≤ k − 1, l ∈ Z. We call domains as in part 2 of Proposition 6.2.1.1 connected on spirals. In particular, this proposition shows that for every D connected on spirals, we can find a connected T k -homogeneous domain that relates to D by (6.2.1.3). Let us give some examples. The domain D = T2P ∩ {|x − P/2| < P/4} is not connected on spirals, whereas D = T2P ∩ {|y| < π/4} is. It follows that D ∩ D is not connected on spirals whereas D ∪ D is. The situation can be more complicated. Set x (x, y, α) := x cos α + y sin α; y  (x, y, α) := −x sin α + y cos α; 0 ≤ α < π/4; P1 := (1/2) |x (P, 2π, −α)|; P2 := (1/2) |y  (P, 2π, −α)|. Then R = {z  = x + i y  : −P1 < x < P1 ; −P2 < y  < P2 } is a fundamental rectangle for T2P in the corresponding coordinates. Set f (y  ) := (P2 − y  )−1 − (y  + P2 )−1 and D0,0 := {z  : −P2 < y  < P2 ; f (y  ) < x < f (y  ) + d} where 0 < d < P1 . Then the domains Dl,m := D0,0 + 2P1 l + 2P2 mi , l, m ∈ Z are ˆ ⊂ T2 . This D ˆ is determined disjoint, and their union D determines a domain D P completely by the intersection of D with the rectangle R = (0, P ) × (−π, π). The ˆ is not connected on spirals. domain D

6.2. A generalization of ρ-trigonometric convexity

225

One more example. Consider the family of lines Ll := {z = x + iy : y = 2 π/(kP )x + lπ/k, x ∈ R}, l ∈ Z. It determines a closed curve (spiral) γ on √TP with 2 n1 = k. The open set Dk = {z : |z − ζ| < , ζ ∈ Ll , l ∈ Z}, 0 < < P/2 π + k 2 , ˆ k on T2 that is connected on spirals, and such that φ−1 (D ˆ k) determines a domain D P k consists of k components, every one of them T -homogeneous. Since the function v in (6.2.1.1) is subharmonic, the function q of (6.2.1.2) is upper semicontinuous and in the D topology on T2P satisfies the inequality Lρ q ≥ 0, where ∂ + ρ2 . (6.2.1.4) Lρ := Δ + 2ρ ∂x Such functions q are called subfunctions with respect to Lρ , or Lρ -subfunctions. Lρ q is a positive measure on T2P . The operator Lρ arises naturally by changing variables z → log z in the Laplace operator Δζ . Exercise 6.2.1.1 Check this. Set ζ = ez . Let us note that if q depends only on the variable y, it is a 2π-periodic ρtrigonometric convex function because Lρ turns into Tρ = (•) + ρ2 (•) (cf. Section 3.2.3). 6.2.2 Consider the solution of the homogeneous boundary problem Lρ q = 0 in D;  q ∂D = 0,

(6.2.2.1)

where D is a domain in T2P and q is bounded in a neighborhood of ∂D with boundary value zero quasi-everywhere. This is a spectral problem for a pencil of differential operators ([Ma]). A solution of this problem can be defined for an arbitrary domain D ⊂ T2P with a boundary of positive capacity. The spectrum of the problem (6.2.2.1) consists of those (complex) ρ for which (6.2.2.1) holds for some function q ≡ 0. The minimal positive point of the spectrum ρ(D) exists iff the spectrum exists. The spectrum exists iff the domain D is connected on spirals. In this case ρ(D) is the order of the minimal harmonic function in every one of the domains Gi that corresponds to D by Proposition 6.2.1.1. The quantity ρ(D) is strictly monotonic. It means that if two domains D1 , D2 ∈ T2P are such that D1 ⊂ D2 and the capacity of D2 \ D1 is positive, then ρ(D2 ) < ρ(D1 ). For example, this is the case of D2 = {|y| < d, d < 2π} and D1 is the same strip without the segment {it : 0 ≤ t ≤ d}. In connection with the multiplicator problem we considered the maximal subharmonic minorant of a function m = H − v where v is a T -automorphic function. From Theorem 6.1.1.2, 2 we can obtain that if v is a T -automorphic function, then GH v is also T -automorphic.

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Exercise 6.2.2.1 Check this. Thus for this case, finding DH in Theorem 6.1.1.1 is reduced to finding a maximal Lρ -subfunction q that satisfies the inequality q(z) ≤ m(z) := [H(ez ) − v(ez )]e−ρx , z ∈ T2P .

(6.2.2.2)

We say that m(z) has an Lρ -subminorant. The idea of ρ(D) gives a possibility for Theorem 6.2.2.1 If m has a non-zero Lρ -subminorant, then ρ(D) ≤ ρ for some component D of the open set M+ := {z : m(z) > 0}. Conversely, if ρ(D) < ρ (strict inequality) for some component D of the set M+ , and m(z) ≥ 0 for all z ∈ T2P , then m has a non-zero Lρ -subminorant. Exercise 6.2.2.2 Prove that M+ is open. 6.2.3 If ρ ∈ / Z, the operator Lρ has a fundamental solution Eρ (• − ζ) in T2P , where ζ is a shift by the torus, i.e., by the modulus P + i2π. It means that Lρ Eρ (• − ζ) = δζ , in D (T2P ), where δζ is the Dirac function, concentrated at ζ. If ρ ∈ Z, there exists, as for operator Tρ and a spherical operator (see Theorem 3.2.4.2, Theorem 3.2.6.3), a generalized fundamental solution Eρ that satisfies the equation Lρ Eρ (• − ζ) = δζ − cos ρ(y − η), ζ = ξ + iη in D (T2P ). Theorem 6.2.3.1 Let ρ > 0, ρ ∈ / Z. Then every Lρ -subfunction on T2P can be represented in the form  Eρ (z − ζ)ν(dζ), (6.2.3.1) q(z) = T2P

where ν = Lρ q. This theorem is the counterpart of Theorems 3.2.3.3, 3.2.6.2. Theorem 6.2.3.2 Let ρ > 0, ρ ∈ Z. Then the mass distribution ν = Lρ v satisfies the condition  e±iρy ν(dz) = 0, (6.2.3.2) T2P

and the representation q(z) = (Ce

 iρy

)+ T2P

Eρ (z − ζ)ν(dζ)

holds with C that is a complex scalar. This theorem is the counterpart of Theorems 3.2.4.2, 3.2.6.2.

(6.2.3.3)

6.2. A generalization of ρ-trigonometric convexity

227

Let D ⊂ T2P and ρ(D) > ρ. Then the operator Lρ has in D the Green function −Gρ (z, ζ, D). Thus for every q that is an Lρ -subfunction in D and bounded from above in D, we have the representation  q(z) = g(z) − Gρ (z, ζ, D)ν(dζ), (6.2.3.4) D

in which ν = Lρ q and g is the minimal majorant on ∂D of the function q, satisfying Lρ g = 0 in D. This is the counterpart of Theorem 2.6.4.3 (F. Riesz representation) and Theorem 3.2.5.1. From (6.2.3.4) one can easily obtain Theorem 6.2.3.3. (Maximum principle) If ρ(D) > ρ and q(z) is an Lρ -subfunction such that q(z) ≤ 0, z ∈ ∂D, then q(z) ≤ 0, z ∈ D. Exercise 6.2.3.1 Prove this. Theorem 6.2.3.4 An Lρ -subfunction in T2P can not attain zero maximum if it is not zero identically. Exercise 6.2.3.1 Prove this by exploiting (6.2.1.2) and properties of subharmonic functions. Theorem 6.2.3.5 Let q be an Lρ -subfunction in T2P . If q(z) ≤ 0 for z ∈ T2P then q(z) ≡ 0. Exercise 6.2.3.2 Prove this using Theorem 3.1.4.7 (**Liouville). Proposition 6.2.3.6 Let qD be the solution of the problem (6.2.2.1) in a domain D with a smooth boundary, corresponding to ρ = ρ(D). Suppose that qD (z0 ) = 1 for some z0 ∈ D. Then ∂qD > 0, ∀z ∈ ∂D. ∂n Exercise 6.2.3.3 Prove this, using properties of positive harmonic functions. 6.2.4 In the part devoted to completeness of an exponential system (Section 6.3) we will need the notion of minimality of a subharmonic function from U [ρ]. A function v ∈ U [ρ] is called minimal if the function v − rρ has no subharmonic minorant for arbitrarily small > 0. If v is T -automorphic, the corresponding Lρ subfunction q is called minimal if the function q − has no Lρ -subminorant in T2P . We formulate one sufficient condition for minimality and one sufficient condition for nonminimality.

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Chapter 6. Application to the Completeness of Exponential Systems

Theorem 6.2.4.1 Let Hρ (q) be the maximal open set on which Lρ q = 0. If there exists a connected component M ⊂ Hρ (q) such that ρ(M ) < ρ, then q is a minimal Lρ -subfunction. For example, q ≡ 0 is minimal. Proposition 6.2.4.2 The function q is nonminimal if q(z) ≥ c or Lρ q − c > 0 for some positive c for all z ∈ T2P . For example, q ≡ c > 0 is nonminimal.

6.3 Completeness of exponential systems in convex domains 6.3.1 Let Λ := {λk }, k = 1, 2, . . . be a set of points in the complex plane C, satisfying the condition λk = 0 and λj = λk , if k = j. Consider the canonical product  (1 − λ/λk ) exp λ/λk . (6.3.1.1) ΦΛ (λ) := k

We suppose in this section that ΦΛ (λ) is an entire function of order 1 and normal type, i.e., a function of exponential type (see [Le, Ch. 1, § 20]. This fact can be expressed in terms of Λ by using the Brelot-Lindel¨of Theorem 2.9.4.2. Exercise 6.3.1.1 Formulate this theorem for entire functions of order 1 and normal type under assumption that ρ(r) ≡ 1. We will suppose that the upper density of zeros (see Section 2.8, Section 5.1). ΔΛ > 0. 6.3.2 Let G ⊂ C be a convex bounded domain containing zero. This last requirement does not restrict any of the further considerations connected to completeness, because exp Λ := {eλj z : λj ∈ Λ} can be replaced by the system {eλj (z−z0 ) : λj ∈ Λ} and eλj (z−z0 ) = Cj eλj z . Let A(G) be the space of holomorphic functions in G with the topology of uniform convergence on compact sets. We will study the completeness of the exponential systems exp Λ := {eλj z : λj ∈ Λ} in A(G). We will be interested in the following questions: 1. completeness of exp Λ in A(G); 2. maximality of G for exp Λ, which is complete in A(G); 3. extremal overcompleteness of exp Λ in A(G) for a maximal G.

(6.3.2.1)

6.3. Completeness of exponential systems in convex domains

229

Let us give precise definitions of maximality and extremal overcompleteness. The completeness means that every function f ∈ A(G) can be approximated on every compact set K  G with arbitrary precision by linear combinations of functions from exp Λ. A convex domain G is called maximal for a system exp Λ, which is complete in A(G) if for every domain G1 such that G  G1 exp Λ is not complete in A(G1 ). A system exp Λ is called extremely overcomplete in A(G) for a maximal G, if for every sequence Λ1 := {λ1j } such that Λ1 ∩ Λ = ∅ and ΔΛ1 > 0 the domain G is not maximal for the system exp Λ ∪ Λ1 . In other words, every essential enlargement of an extremely overcomplete system enlarges also the maximal domain of completeness. 6.3.3 Let

hΛ (φ) := lim sup log |ΦΛ (reiφ )|r−1 r→∞

be the indicator of ΦΛ . It is a 1-trigonometrically convex function or simply a trigonometrically convex function (t.c.f). Let GΛ be the conjugate indicator diagram of ΦΛ , i.e., a convex domain of the form GΛ := {z : max (zeiφ ) ≤ hΛ (φ)}. z∈GΛ

Let us describe conditions for completeness, maximality and extremal overcompleteness when Λ is a regular set (see Section 5.6) and ΦΛ is a CRG-function (see Section 5.6). We say that GΛ is enclosed in G if it can be enclosed in G by parallel translation, enclosed with sliding, if it can be moved after enclosing only in one direction, enclosed rigidly if it is impossible to move after enclosing, freely enclosed in every other case of enclosing. Theorem 6.3.3.1 Let Λ be a regular set. Then the following holds: 1. {exp Λ is not complete in A(G)} ⇐⇒ {GΛ is freely enclosed in G}; 2. {G is maximal for exp Λ} ⇐⇒ {GΛ is not freely enclosed in G}; 3. {exp Λ is extremely overcomplete in A(G)} ⇐⇒ {GΛ is enclosed rigidly in G}. Let us note that G is maximal for exp Λ but not extremely overcomplete if and only if GΛ is enclosed with sliding in G. This theorem is a corollary of the more general Theorem 6.3.4.1, but will be proved independently in Section 6.3.10. 6.3.4 If Λ is not regular, it is natural to exploit the notion of a limit set (see Section 3.1) to characterize exp Λ. Suppose the limit set of ΦΛ has the form Fr[ΦΛ ] := {v(λ) = |λ|(ch1 + (1 − c)h2 )(arg λ) : c ∈ [0; 1]} where h1 , h2 are t.c.f.

230

Chapter 6. Application to the Completeness of Exponential Systems

Such a limit set is a particular case of Uind (6.1.2.2). It is called an indicator limit set and it is indeed a limit set of an entire function (see Exercise 6.1.2.4). The asymptotic behavior of the set Λ (i.e., the limit set of the corresponding mass distribution) can be described completely using Theorem 3.1.5.2. Exercise 6.3.4.2 Do that. We will call such Λ an indicator set. Denote by G1 , G2 the conjugate diagram of h1 , h2 . Since G1 , G2 are convex, the set αG1 + βG2 := {αz1 + βz2 : z1 ∈ G1 , z2 ∈ G2 }, α, β > 0 is also convex and is a conjugate diagram of the t.c.f. h := αh1 + βh2 . Theorem 6.3.4.1 Let a set Λ be an indicator set. Then the following holds: 1. {exp Λ is not complete in A(G)} ⇐⇒ {G1 and G2 are freely enclosed in G}; 2. {G is maximal for exp Λ} ⇐⇒ {G1 and G2 are enclosed in G and at least one of them is not freely enclosed in G}; 3. {exp Λ is extremely overcomplete in A(G)} ⇐⇒ {cG1 + (1 − c)G2 is enclosed rigidly in G ∀c ∈ [0; 1]}. This theorem is proved in Section 6.3.10. The equality holds: hΛ = max(h1 , h2 ).

(6.3.4.1)

Thus the conjugate diagram GΛ of the function hΛ is the convex hull of G1 and G2 . Let us note that the indicator hΛ does not determine the completeness of the system exp Λ if Λ is not a regular set, as the following example shows. Example 6.3.4.1 Let G1 := {z = x + iy : x = 1; −1 ≤ y ≤ 1}, G2 := {z = x + iy : x = −1; −1 ≤ y ≤ 1}, and

G = {z : |z| < 1 + }

with a small . Exercise 6.3.4.3 Prove that G1 and G2 are freely enclosed in G and their convex hull is not enclosed. Let Λ be a set such that the interior of GΛ coincides with G. If Λ is a regular set, then exp Λ is complete in A(G), G is maximal for exp Λ and exp Λ is extremely overcomplete in A(G). If Λ is an indicator set, then the first two assertions hold but exp Λ can be not extremely overcomplete:

6.3. Completeness of exponential systems in convex domains

231

Example 6.3.4.2 Set G1 := {z = x + iy : −1 ≤ x ≤ 0; y = 0}; G2 := {z = x + iy : x = 1; −1 ≤ y ≤ 1}. Here GΛ is a triangle in which G1 is freely enclosed and G2 is rigidly enclosed, but cG1 + (1 − c)G2 is free enclosed for all c : 0 < c < 1. Exercise 6.3.4.4 Check this. Example 6.3.4.3 Set G1 := {z = x + iy : x = −1; y ∈ [−1; 1]}; G2 := {z = x + iy : x = 1; y ∈ [−1; 1]}. Exercise 6.3.4.5 Check that G1 and G2 are enclosed with sliding in GΛ . If G1 and G2 are rigidly enclosed in G it does not imply in general that cG1 + (1 − c)G2 are rigidly enclosed for all c ∈ [0; 1]. Example 6.3.4.4 Let G1 be an equilateral triangle inscribed in the circle |z| = 1, let G2 be the same triangle rotated by the angle π/6, and let G be the unit disc. Exercise 6.3.4.6 Show that 12 (G1 + G2 ) is freely enclosed in G. If G1 , G2 ⊂ G and G1 ∩ G2 is rigidly enclosed in G, then cG1 + (1 − c)G2 is rigidly enclosed for c ∈ [0; 1]. Exercise 6.3.4.7 Check this. However this is not a necessary condition. Example 6.3.4.5 G := {z = x + iy : |x| < 1; |y| < 1}; G1 := {z = x + iy : x ∈ (−1, 1); −x > y > −1}; G2 := {z = x + iy : x ∈ (−1, 1); −1 < y < x}. Exercise 6.3.4.8 Check that every triangle cG1 + (1 − c)G2 is rigidly enclosed in G and G1 ∩ G2 is freely enclosed. 6.3.5 Consider in more detail the conditions for extremal overcompleteness in the case when Λ is an indicator set and GΛ = G or, in other words, if hΛ = hG .

(6.3.5.1)

We can suppose that h1 and h2 are linearly independent, otherwise we exploit Theorem 6.3.3.1. If, for example, the inequality h1 (φ) ≤ h2 (φ), ∀φ, holds, the extremal overcompleteness is in the case when G1 is rigidly enclosed in G2 because G1 ∩ G2 = G1 , and this case was mentioned above (Exercise 6.3.4.7).

232

Chapter 6. Application to the Completeness of Exponential Systems

Consider the general case. Denote g(φ) := |h1 − h2 |(φ), and set ΘΛ := {φ : g(φ) > 0}. This is an open set on the unit circle. Denote as IΛ := (α1 , α2 ) the maximal interval contained in ΘΛ and denote by dΛ its length. Since g(φ) is continuous, g(αj ) = 0, j = 1, 2. (6.3.5.2) If also at least one of the conditions lim inf φ∈IΛ ,φ→αj

g(φ) = 0, j = 1, 2, φ − αj

is fulfilled, we say g is zero with tangency on ∂IΛ . Theorem 6.3.5.1 Suppose Λ is an indicator set that satisfies (6.3.5.1). In order that exp Λ be extremely overcomplete in A(G) it is necessary and sufficient that at least one of the following conditions holds: 1. dΛ < π; 2. dΛ = π and g is zero with tangency on ∂IΛ . This theorem is proved in Section 6.3.11. 6.3.6 We call Λ periodic if Fr[ΦΛ ] is a periodic limit set (see Theorem 4.1.7.1). In such a case all the limit set is determined by one subharmonic function v ∈ U [1] (see (4.1.3.1)). Let us characterize the system exp Λ for periodic Λ. Set hG (φ) := max{(zeiφ ) : z ∈ G}, m(λ, G, v) := |λ|hG (arg λ) − v(λ).

(6.3.6.1) (6.3.6.2)

Denote by GG v the maximal subharmonic minorant of the function m(λ,G,v). A function w ∈ U [1] is called minimal if the function w − |λ| has no subharmonic minorant in U [1] for every small > 0. The harmonic function of the form H(λ) := |λ|(A cos(arg λ) + B sin(arg λ)),

(6.3.6.3)

for example, is minimal. We will denote as HARM the set of functions of the form (6.3.6.3). Theorem 6.3.6.1 Let Λ be a periodic set. The following holds: 1. {exp Λ is not complete in A(G)} ⇐⇒ {GG v exists and is non-minimal}; 2. {G is maximal for exp Λ} ⇐⇒ {GG v exists and is minimal}; 3. {exp Λ is extremely overcomplete in A(G)} ⇐⇒ {GG v ∈ HARM}. This theorem is proved in Section 6.3.12.

6.3. Completeness of exponential systems in convex domains

233

6.3.7 Let us characterize the completeness of expΛ for periodic Λ in other terms. For this we need the information that was presented in Section 6.2. We will take ρ = 1. Denote qΛ (z) := vΛ (ez )e−x (6.3.7.1) (compare with (6.2.1.2)). As it was explained in Section 6.2 this function is an L1 -subfunction on the torus T2P . Set m(z, G, qΛ ) = hG (y) − qΛ , D(G, Λ) := {z : m(z, G, qΛ ) > 0} ⊂ T2P . The set D(G, Λ) is open because −m is an upper semicontinuous function (see Theorem 2.1.2.4), denote ρ(Λ, G) := min ρ(M ) where the minimum is taken over all components M of D(G, Λ), and it is attained on one of the components because they are not intersecting and T2P is compact. Exercise 6.3.7.1 Explain this in detail, using properties of ρ(D) (Section 6.2). Theorem 6.3.7.1 If ρ(Λ, G) ≥ 1,

(6.3.7.2)

then exp Λ is complete in G. This theorem is proved in Section 6.3.12. Let w := gG qΛ (z) be the maximal L1 -subminorant of m(z, G, qΛ ). Denote by HΛ the open set in T2P where L1 w = 0. Theorem 6.3.7.2 If there exists a component M of HΛ such that ρ(M ) < 1, then w is minimal, and, hence, G is maximal for exp Λ. This theorem follows directly from Theorem 6.2.4.1. It is not known if the condition (6.3.7.2) is necessary. Consider in detail the situation in which the domain G coincides with GΛ , the conjugated indicator diagram of hΛ , i.e., we suppose that hG (φ) = hΛ (φ), ∀φ.

(6.3.7.3)

In this case m(z, G, qΛ ) ≥ 0 and we obtain the following criterion: Theorem 6.3.7.3 In order that exp Λ be complete in GΛ it is necessary and sufficient that ρ(Λ, GΛ ) ≥ 1. (6.3.7.4) This theorem is proved in Section 6.3.12. The condition (6.3.7.3) automatically implies maximality if there is completeness. Since (6.3.7.5) hΛ (y) = max{qΛ (x + iy) : x ∈ [0; P ]}, the function m(z, GΛ , qΛ ) has a zero in x for every fixed y. Thus the set D(G, Λ) does not contain any curve y = const on the torus.

234

Chapter 6. Application to the Completeness of Exponential Systems

Theorem 6.3.7.4 Let G0 be a strictly convex domain and let D0 ⊂ T2P be such that T2P \ D0 intersect every line {y = y0 }, y0 ∈ [0, 2π]. Then there exists a periodic Λ such that GΛ = G0 , D(GΛ , Λ) = D0 .

(6.3.7.6)

This theorem is proved in Section 6.3.13. Example 6.3.7.1 Let D0 be the complement in T2P to the set M := {z = x + iy : x = f (y), y ∈ [0; 2π]}

(6.3.7.7)

where f (y) is a continuous 2π-periodic function satisfying the condition 0 < f (y) < P. Then ρ(D0 ) = ∞, because this domain is not connected on spirals (see Section 6.2.). It means that for every strictly convex G0 there exists a periodic Λ such that GΛ = G0 and exp Λ is extremely overcomplete in G0 . Example 6.3.7.2 Let D0 be the complement to the set   P y, 0 ≤ y ≤ 2π . M := z = x + iy : x = 2π Then ρ(D0 ) =

 1 1 + (2π/P )2 2

(6.3.7.8)

(see Section 6.3.13). Thus, choosing P , and using Theorem 6.3.7.4, it is possible make exp Λ complete or non-complete in G0 (= GΛ ) for every strictly convex domain G0 . 6.3.8 Now pass to generalizations. Denote by DG the natural domain of definition of the operation GG , i.e., the set of v ∈ U [1] for which m(λ, G, v) (see (6.3.6.2)) has a subharmonic minorant belonging to U [1]. Let ΦΛ be defined by the equality (6.3.1.1). The condition that for every v ∈ Fr[ΦΛ ] the function m(λ, G, v) has a subharmonic minorant belonging to U [1] is possible to express by the relation Fr[ΦΛ ] ⊂ DG

(6.3.8.1)

(compare with (6.1.1.1)). We call the set U ⊂ U [1] minimal (U ∈ MIN) if for arbitrarily small > 0 there exists w = w ∈ U such that the function w − |λ| has no subharmonic minorant, belonging to U [1]. Let us note that if U contains a minimal function (in the sense of Section 6.3.6), then U ∈ MIN. We denote the image of Fr[ΦΛ ] under the mapping by the operator GG as JG (Λ).

6.3. Completeness of exponential systems in convex domains

235

Theorem 6.3.8.1 The following holds: 1. {expΛ is not complete in A(G)} ⇐⇒ {((6.3.8.1) holds) ∧ (JG (Λ) ∈ / MIN)}; 2. {G is maximal for expΛ } ⇐⇒ {((6.3.8.1) holds) ∧ (JG (Λ) ∈ MIN)}; 3. {exp Λ is extremely overcomplete for maximal G} ⇐⇒ {((6.3.8.1) holds) ∧ (JG (Λ) ∈ HARM)}; 6.3.9 In the proof of Theorem 6.3.8.1 that we are going to prove now we exploit Theorem 6.3.9.1. (A.I. Markushevich) see [Le, Ch. 4, § 7] Let A(C \ G) be a class of functions ψ which are holomorphic in C \ G and equal to zero in infinity. In order that the system exp Λ be complete in A(G), it is necessary and sufficient that the function  eλz ψ(z)dz,

Φ(λ) :=

(6.3.9.1)

Lψ

where ψ ∈ A(C \ G), and Lψ  G is a rectifiable closed curve, has the following property: the condition (6.3.9.2) Φ(λk ) = 0, ∀λk ∈ Λ implies Φ(λ) ≡ 0. Proof of Theorem 6.3.8.1, 1. Necessity. Let exp Λ be not complete . By Theorem 6.3.9.1 Φ(λk ) = 0, but Φ(λ) ≡ 0. The function g(λ) := Φ(λ)/ΦΛ (λ), where ΦΛ is from (6.3.1.1), is an entire function and it has order one and normal or minimal type by Theorem 2.9.3.1. Set ug := log |g(λ)|; uΦ (λ) := log |Φ(λ)|; uΛ (λ) := log |Φ(λ)|. We have from (6.3.9.1) uΦ (λ) ≤ max{(λz) : z ∈ Lψ } + Cψ }, where Cψ is a constant, depending possibly on ψ. This implies that uΦ (λ) ≤ hG1 (φ)r + Cψ , λ = reiφ ,

(6.3.9.3)

for some convex domain G1  G. Let v ∈ Fr[ΦΛ ]. Choose a sequence tj → ∞ for which (uΛ )tj → v, and the sequences (uΦ )tj and (ug )tj also converge to v Φ and v g respectively. From the equality ug (λ) = uΦ (λ) − uΛ (λ) we obtain v g (λ) = v Φ (λ) − v(λ) where v g ∈ Fr[g], v Φ ∈ Fr[Φ]. Since (6.3.9.3) implies v Φ (λ) ≤ hG1 (φ)r, v g (λ) ≤ hG1 (φ)r − v(λ)

(6.3.9.4)

and it means that for every v ∈ Fr[ΦΛ ] GG1 v and hence GG v exist, i.e., the condition (6.3.8.1) holds.

236

Chapter 6. Application to the Completeness of Exponential Systems

Let us show that the condition JG (Λ) ∈ / MIN is satisfied. We have for some δ > 0 the relation hG1 (φ) − hG (φ) ≤ −δ. From (6.3.9.4) we obtain v g (λ) + δr ≤ m(λ, G, v).

(6.3.9.5)

The left-hand side of the inequality (6.3.9.5) belongs to U [1]. Thus wv := GG1 v satisfies the condition v g (λ) + δr ≤ wv (λ) for every v ∈ Fr[ΦΛ ]. It means that / MIN. JG (Λ) ∈ Necessity is proved.  For proving sufficiency we exploit the following assertion. Theorem 6.3.9.2 (I.F. Krasichkov-Ternovskii) Suppose there exists an entire function g such that (6.3.9.6) hgΦΛ (φ) < hG (φ), ∀φ. Then the system exp Λ is not complete for some convex domain G1  G. This theorem connects the problem of completeness to the multiplicator problem. Proof of Theorem 6.3.9.2. Let g(λ) satisfy (6.3.9.6). Denote by ψ(z) the Borel transformation for Φ(λ) := g(λ)ΦΛ (λ). By the P´ olya Theorem (see, for example, [Le, Ch. 1, § 20]), all the singularities of ψ are contained in a convex domain GΦ which is the conjugate diagram of the indicator hΦ (φ). Thus the representation (6.3.9.1) holds with Lψ that embraces GΦ . It follows from (6.3.9.6) that GΦ  G. Thus it is possible to choose Lψ between ∂GΦ and ∂G. Since (6.3.9.2) for Φ is fulfilled and Φ(λ) ≡ 0, exp Λ is non-complete in some convex G1  G such that  Lψ  G1 by Theorem 6.3.9.1. Now we can prove sufficiency in Theorem 6.3.8.1, 1. From the condition JG (Λ) ∈ / MIN it follows that one can choose δ > 0 such that ∀v ∈ Fr[ΦΛ ] the functions wv − δr where wv := GG , have subharmonic minorants. As we already said in Section 6.3.2, completeness does not depend on shift by any fixed z0 . Thus we can suppose that 0 ∈ G and, hence, hG (φ) > 0 for all φ. Let γ < 2δ be such that hG (φ) − γ > 0 and G1  G satisfy hG1 (φ) − γ/3 > 0, hG (φ) − hG1 (φ) ≤ γ/2.

(6.3.9.7)

DG1 ⊃ Fr[ΦΛ ],

(6.3.9.8)

Let us check that Indeed, for v ∈ Fr[ΦΛ ] we have m(λ, G1 , v) := hG1 (φ)r − v(λ) ≥ hG (φ)r − (γ/2)r − v(λ) ≥ hG (φ)r − v(λ) − δr ≥ wv − δr.

(6.3.9.9)

6.3. Completeness of exponential systems in convex domains

237

Since the right-hand side of (6.3.9.9) has a subharmonic minorant from U [1], then (6.3.9.8) is proved. By Theorem 6.1.1.1 there exists a multiplicator g(z) ∈ A(1) such that (6.3.9.10) hgΦ (φ) ≤ hG1 (φ) < hG (φ). From Theorem 6.3.9.2 we obtain that exp Λ is non-complete in G. Proof of Theorem 6.3.8.1, 2. Necessity. Let Gj , j = 1, 2, . . . be a sequence of convex domains, satisfying the conditions Gj  G, Gj ↓ G. Since exp Λ is noncomplete in every A(Gj ), DGj ⊃ Fr[ΦΛ ] by Theorem 6.3.8.1, 1. The sequence wj := GGj v satisfies wj (λ) ≤ hGj (φ)r − v(λ), λ ∈ C. Since {wj } is compact and hGj → hG , one can find a subsequence with the limit w ∈ U [1]. Then w(λ) ≤ hG (φ)r − v(λ). Hence GG v exists. If JG ∈ MIN would not hold, then, by Theorem 6.3.8.1, 1, exp Λ is noncomplete in A(G), which contradicts maximality. Necessity is proved. Let us prove sufficiency. Completeness of exp Λ in A(G) follows from Theorem 6.3.8.1, 1. We will prove that expΛ is non-complete in A(G1 ) for every G1  G under the condition DG ⊃ Fr[ΦΛ ]. Set δ := min[hG1 (φ) − hG (φ)] > 0. φ

Then ∀v ∈ Fr[ΦΛ ], GG v + δr ≤ hG1 (φ)r − v(λ), λ ∈ C. This means that GG1 v ≥ GG v + δr. Hence JG1 (Λ) ∈ / MIN and, by Theorem 6.3.8.1,  1, exp Λ is non-complete in A(G1 ). Proof of Theorem 6.3.8.1, 3. Necessity. By Theorem 6.3.8.1, 2 from maximality G (6.3.8.1) follows. We will prove that GG v ∈ HARM ∀v ∈ Fr[ΦΛ ]. Suppose it is not fulfilled, i.e., there exists v0 ∈ Fr[ΦΛ ] such that the mass distribution ν0 of the function w0 = GG v0 is not zero. By Proposition 6.1.1.3 there exists a multiplicator g such that v0 + w0 ∈ Fr[gΦΛ ]. Let Λ0 be the set of zeros of g. Since ν0 ∈ FrΛ0 , Δ(Λ0 ) > 0, because ν0 = 0 and by the definitions in Section 3.3.1. We can shift a little zeros of g and suppose without lack of generality that they are simple and Λ0 ∩ Λ = ∅. The condition for a multiplicator gives the inequality: hgΦΛ (φ) ≤ hG (φ), ∀φ. It implies m(λ, G, vΠ ) = rhG (φ) − vΠ ≥ 0

238

Chapter 6. Application to the Completeness of Exponential Systems

for all vΠ ∈ Fr[gΦΛ ]. It means that m(λ, G, vΠ ) has zero as a minorant ∀vΠ ∈ Fr[gΦΛ ], i.e., DG ⊃ Fr[gΦΛ ]. So the domain G is maximal although the system exp Λ is replaced with the system exp(Λ ∪ Λ0 ). This contradicts the extremal overcompleteness. Hence, ν0 ≡ 0 and w0 = GG v0 ∈ HARM. Necessity is proved. Let us prove sufficiency. Let the condition GG v ∈ HARM ∀v ∈ Fr[ΦΛ ] hold. Suppose that there exists Λ0 such that ΔΛ0 > 0 and G is maximal for the system exp(Λ ∪ Λ0 ). Theorem 6.3.8.1, 2 implies DG ⊃ Fr[ΦΛ1 ],

(6.3.9.11)

where Λ1 = Λ ∪ Λ0 . For every v0 ∈ Fr[ΦΛ0 ] one can find v ∈ Fr[ΦΛ ] such that v1 := v0 + v ∈ Fr[ΦΛ1 ]. The condition ΔΛ0 > 0 implies that one can choose v0 for which the Riesz measure ν0 ≡ 0. For w1 = GG v1 one has the inequality w1 ≤ rhG − v1 by (6.3.9.11), so w1 + v0 ≤ rhG − v holds. Hence wv := GG v satisfies the inequality (w1 + v0 )(λ) ≤ wv (λ), ∀λ ∈ C.

(6.3.9.12)

Let us show that (6.3.9.12) is impossible.Indeed, since wv ∈ HARM w := w1 + v0 − wv ≤ 0 and w ∈ U [ρ]. Thus w ≡ 0. However the Riesz measure νw ≥ ν0 ≡ 0, hence w ≡ 0. This contradiction proves sufficiency.  6.3.10 Now we prove Theorems 6.3.3.1, 6.3.4.1 and 6.3.5.1. We need some auxiliary assertions. Lemma 6.3.10.1 Let v := rh1 (φ) and G1 be the conjugated diagram of h1 . Then the following holds: 1. 2. 3. 4.

{G1 {G1 {G1 {G1

is is is is

freely enclosed in G} ⇐⇒ {GG v is non-minimal}; enclosed to G but not free enclosed}⇐⇒ {GG v is minimal}; rigidly enclosed in G} ⇐⇒ {GG v ∈ HARM}; not enclosed in G} ⇐⇒ {GG v does not exist}.

To prove this lemma we need the following two lemmas. Lemma 6.3.10.2 Let v := rh(φ). Then GG v = rh1 (φ) where h1 is the maximal trigonometrically convex minorant of the function m(φ, G, h) := hG (φ) − h(φ).

6.3. Completeness of exponential systems in convex domains

239

Proof. Let v1 = GG v. Since v[t] = v for all t > 0, (v1 )[t] = GG v[t] = GG v by Theorem 6.1.1.2, 2. Thus the function vˆ1 :=

 ∗ sup(v1 )[t] (λ) ≥ v1 (λ) t

and is also a subharmonic minorant belonging to U [1]. Thus v1 = vˆ1 . However, the function vˆ1 is invariant with respect to the transformation (•)[t] . Hence it has the form rh1 (φ). The maximality of h1 (φ) follows from the maximality v1 .  Lemma 6.3.10.3 In order that v := rh1 be a minimal function, it is necessary and sufficient that G1 , the conjugate diagram of h1 , be a segment (in particular, a point). Proof. Let v = rh1 be minimal and let G1 be the conjugate diagram of h1 . If G1 is not the segment, then it contains some disc of radius δ > 0. Hence there exists a trigonometric function A cos(φ − φ0 ) such that δ + A cos(φ − φ0 ) ≤ h1 (φ). Multiplying this inequality by r, we obtain that v − δr has a harmonic (and hence subharmonic) minorant. This contradicts minimality. Inversely, suppose v is not minimal. Then there exists δ > 0 and t.c.f. h2 (φ) such that h2 (φ) ≤ h1 (φ) − δ. (6.3.10.1) For every t.c.f. h2 there exists a trigonometric function A cos(φ − φ0 ) such that h2 (φ) + A cos(φ − φ0 ) ≥ 0.

(6.3.10.2)

This corresponds to a shift of the diagram which contains zero. From (6.3.10.1) and (6.3.10.2) we obtain δ − A cos(φ − φ0 ) ≤ h1 (φ), which means that G1 contains some disc of radius δ > 0. So it is not a segment.



Proof of Lemma 6.3.10.1. G is freely enclosed iff the following assertion holds: there exists δ > 0 and a trigonometrical function A cos(φ − φ0 ) such that the inequality (6.3.10.3) h1 (φ) + δ − A cos(φ − φ0 ) ≤ hG (φ) holds.

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Chapter 6. Application to the Completeness of Exponential Systems

Exercise 6.3.10.1 Prove this. Let GG v be non-minimal. By Lemma 6.3.10.2 it has the form w2 = rh2 , where h2 is the maximal trigonometrically convex minorant of m(φ, G, h1 ). There exists δ > 0 such that the function w2 − δr has the maximal subharmonic minorant v3 = rh3 (φ). Let A cos(φ − φ0 ) be a trigonometric function for which h3 (φ) + A cos(φ − φ0 ) ≥ 0. In addition, h3 (φ) ≤ h2 (φ) − δ, h2 (φ) ≤ hG − h1 (φ). From this we obtain (6.3.10.3) and hence that G1 is free enclosed. Inversely, let G1 be freely enclosed in G. From (6.3.10.3) it follows that δ − A cos(φ − φ0 ) ≤ hG (φ) − h1 (φ).

(6.3.10.4)

Multiplying (6.3.10.4) by r, we obtain that m(λ, G, v) has a minorant v0 = r(δ − A cos(φ − φ0 )) which obviously is non-minimal. Hence, GG v is non-minimal. G1 is enclosed in G with sliding, hence there does not exist δ > 0 such that (6.3.10.3) is fulfilled, but there exists a segment with support function E(φ) = B| sin φ| + A cos(φ − φ0 ), such that the inequality h1 (φ) + E(φ) ≤ hg (φ)

(6.3.10.5)

holds. Exercise 6.3.10.2 Prove this. Let GG v be minimal. By Lemma 6.3.10.2 it has the form w2 = rh2 and by Lemma 6.3.10.3, h2 = E(φ). Thus E(φ) ≤ (hG − h1 )(φ), which is equivalent to (6.3.10.5). Prove 2, suppose G is not freely enclosed and hence only (6.3.10.5) is possible. If GG v were non-minimal, (6.3.10.3) would follow, as it was proved above. This contradicts the supposition. The rigid enclosure is equivalent only to the inequality of the form h(φ) − A cos(φ − φ0 ) ≤ hG (φ) ∀φ, and impossibility of enclosure is equivalent to the impossibility of even such an inequality. Thus all other assertions of the lemma can be proved analogously. Exercise 6.3.10.3 Do this in detail.



Proof of Theorem 6.3.3.1. Regularity of Λ means that Fr[ΦΛ ] = {v0 } where v0 = rhΛ . Thus JG (Λ) = {GG v0 } and all the assertions of Theorem 6.3.3.1 follows from Theorem 6.3.8.1 and Lemma 6.3.10.1.  Exercise 6.3.10.4 Check this in detail.

6.3. Completeness of exponential systems in convex domains

241

For proving Theorem 6.3.4.1 we need an additional Lemma 6.3.10.4 Let Λ be an indicator set, v1 = rh1 , v2 = rh2 . Then / MIN} ⇐⇒ {GG v1 and GG v2 are non-minimal}. {JG (Λ) ∈ Proof. Suppose w1 := GG v1 and w2 := GG v2 are not minimal, i.e., w1 − δr and w1 − δr have subharmonic minorants g1 and g2 . Then cg1 + (1 − c)g2 is a minorant of the function cw1 + (1 − c)w2 − δr, i.e., JG (Λ) ∈ / MIN. The inverse implication is trivial.  Exercise 6.3.10.5 Prove this. Proof of Theorem 6.3.4.1. Suppose exp Λ is not complete. By Theorem 6.3.8.1 JG ∈ / MIN. By Lemma 6.3.10.4, GG v1 and GG v2 are not minimal. Hence G1 and G2 are freely enclosed in G by Lemma 6.3.10.1. Since every one of these assertions is reversible, the inverse implication also holds. Analogously the other cases are proved.  Exercise 6.3.10.6 Prove all this in detail. 6.3.11 To prove Theorem 6.3.5.1 we need some auxiliary assertions. Lemma 6.3.11.1 Let φ0 be a maximum point of t.c.f. h(φ) and h(φ0 ) ≥ 0. Then h(φ) ≥ h(φ0 ) cos(φ − φ0 ), |φ − φ0 | ≤ π/2.

(6.3.11.1)

Proof. We write y(φ) := h(φ0 ) cos(φ − φ0 ). We have y(φ0 ) = h(φ0 ) and y(φ) is a trigonometric function. If y(φ1 ) = h(φ1 ) for some φ1 such that |φ1 − φ0 | < π/2 this contradicts Theorem 3.2.5.2. If y(φ) does not intersect h(φ), this contradicts Theorem 6.2.3.4 applied to the function h(φ) − y(φ), which is an Lρ -subfunction with ρ = 1.  Lemma 6.3.11.2 Let H(φ) be a trigonometric function on the interval I = (α, β) of length ≤ π, such that H(φ) = 0 at one of the ends of I. Then every one of the conditions 1. H(φ0 ) = 0, φ0 ∈ (α; β);

2. H(φ) is zero on ∂I with tangency;

implies H(φ) ≡ 0, φ ∈ I. Exercise 6.3.11.1 Prove this. Lemma 6.3.11.3 Let g ≥ 0 be a continuous periodic function, and let ΘΛ , IΛ , dΛ be defined as in Theorem 6.3.5.1. In order that its maximal t.c.minorant be a trigonometrical function, it is necessary and sufficient satisfying at least one of the conditions: 1. dΛ < π; 2. dΛ = π and g(φ) is zero with tangency on ∂I.

242

Chapter 6. Application to the Completeness of Exponential Systems

Proof. Necessity. Suppose dΛ > π. Without loss of generality we can suppose that IΛ = (α; −α), where α > π/2. Set cos+ φ := max(cos φ, 0),  g(φ) : φ ∈ (−α; α) . (6.3.11.2) a = inf cos+ φ We have a > 0. Set

 a1 cos φ, h(φ) := 0

|φ| ≤ π/2 |φ| > π/2,

(6.3.11.3)

where a1 ≤ a. The function h(φ) is a t.c.minorant of g(φ) and it is not a trigonometric function, which contradicts the supposition. Thus dΛ ≤ π. Suppose dΛ = π and the condition to be zero with tangency on ∂I does not hold. Then for a defined by (6.3.11.2) the condition a > 0 holds and h(φ) defined by (6.3.11.3) is a non-trigonometric minorant of g that contradicts the supposition. Sufficiency. Let the first condition hold and let I = (α; β) be an arbitrary interval belonging to ΘΛ ; let h(φ) be the maximal t.c.minorant of g(φ). Set H(φ) := h(φ0 ) cos(φ − φ0 ), where φ0 is the maximum point of h(φ) on I. From inequality (6.3.11.1) and the conditions g(α) = g(β) = 0 follows H(α) = H(β) = 0. Then, by Lemma 6.3.11.2, we obtain H(φ) ≡ 0. Thus h(φ0 ) = 0 and h(φ) ≡ 0 for φ ∈ (α; β), i.e., h(φ) is trigonometric. Let the second condition be fulfilled. Lemma 6.3.11.1 implies that H(φ) is zero with tangency on ∂I. By Lemma 6.3.11.2 we obtain that h(φ) ≡ 0.  Proof of Theorem 6.3.5.1. Necessity. Let us note that if v ∈ Fr[ΦΛ ], then for c ∈ [0; 1] we have the equality   (6.3.11.4) m(λ, G, v) = r c(h2 − h1 )+ + (1 − c)(h2 − h1 )− (φ) := rm(φ, c). Let exp Λ be extremely overcomplete in A(G). By Theorem 6.3.8.1 JG ⊂ HARM, i.e., for every c ∈ [0; 1] the maximal t.c.minorant of the function m(φ, c) is trigonometric. Since ∀c ∈ [0; 1], ΘΛ = {φ : m(φ, c) > 0} the necessity follows from Lemma 6.3.11.3. Sufficiency. It follows directly from Lemma 6.3.11.3. Exercise 6.3.11.2 Explain this.



6.3. Completeness of exponential systems in convex domains

243

6.3.12 To prove Theorem 6.3.6.1 we need Lemma 6.3.12.1 If w ∈ U [1] is non-minimal, then C(w) := {w[t] : 1 ≤ t ≤ eP } ∈ / MIN . It follows from Theorem 6.1.1.2, 2 Exercise 6.3.12.1 Explain this in detail. Proof of Theorem 6.3.6.1. By Theorem 6.1.1.2, 2. JG (Λ) = C(GG v). Thus Lemma / MIN if and only if GG v is not minimal. Thus Theo6.3.12.1 implies that JG (Λ) ∈ rem 6.3.8.1 implies Theorem 6.3.6.1, 1 and 2. Suppose JG (Λ) ⊂ HARM. Hence, GG v = rH0 (φ), where H0 is trigonometric.  Inversely, Lemma 6.3.12.1 implies JG (Λ) = {rH0 (φ)}. Proof of Theorem 6.3.7.1. Let ρ(Λ, G) > 1. Suppose wq := gG q exists. By definition of ρ(Λ, G) we have wq (z) ≤ 0 for z ∈ ∂D(Λ, G). By Theorem 6.2.3.3 (Maximum principle) wq ≤ 0 for z ∈ D(G, Λ). Also wq ≤ 0 for z ∈ T2P \ D(G, Λ) by definition of D(Λ, G). By Theorem 6.2.3.4 wq ≡ 0 and hence is minimal. So exp Λ is complete by Theorem 6.3.6.1. If ρ(Λ, G) = 1, then the system exp Λ is complete for every Gn  G, because of strict monotonicity of ρ(•) (see Section 6.2.2) so G is the maximal domain.  For the proof of Theorem 6.3.7.3 we need an auxiliary assertion. We suppose that D is an image on T2P by the map (6.2.1.3) of the domain G with a smooth boundary. Theorem 6.3.12.2 Let D ⊂ T2P and ρ(D) ≤ 1. Then ρ(T2P \ D) > 1, if DΛ = {z > 0}. For the proof we need the following assertion which was proved originally by A. Eremenko and M. Sodin: Theorem 6.3.12.3 (Eremenko, Sodin) Let Γ be a Jordan curve, connecting 0 and ∞, T Γ = Γ for some T > 1. Let D+ , D− be domains, into which Γ divides the plane, and let ρ1 , ρ2 be the orders of the minimal harmonic functions in D+ and D− respectively. Then 1 1 + ≤ 2, ρ1 ρ2 and equality is attained only if Γ consists of two rays. We will give prove this theorem in Section 6.3.14. Proof of Theorem 6.3.12.2. Let ρ1 = ρ(D), and suppose q1 (z) is a solution of boundary problem (6.2.2.1), ρ2 = ρ(T2P \ D), q2 (z) is a solution of the corresponding boundary problem. Then the image of the boundary under the map λ = ez (we denote it as Γ) satisfies the conditions of Theorem 6.3.12.3 and the functions

244

Chapter 6. Application to the Completeness of Exponential Systems

v1 (λ) := q1 (log λ)|λ|ρ1 and v2 (λ) := q2 (log λ)|λ|ρ2 are positive harmonic functions in D+ , D− with orders ρ1 and ρ2 respectively. By Theorem 6.3.12.3 we obtain 1/ρ(D) + 1/ρ(T2P \ D) ≤ 2. and equality holds only if Γ is a pair of rays, i.e., DΛ = {z > 0}.



Proof of Theorem 6.3.7.3. Necessity. Suppose ρ(Λ, GΛ ) < 1. Let us prove that exp Λ is not complete. To this end we construct an L1 -minorant of m(z, GΛ , Λ) and prove that it is not minimal. Let D0  D(GΛ , Λ) be a domain with smooth boundary for which ρ(D0 ) = 1. This is possible because of strict monotonicity ρ(D) (Section 6.2.2). Let q0 be a solution of the problem (6.2.2.1) satisfying the condition 0 < max{q0 (z) : z ∈ D0 } ≤ min{m(z, GΛ , Λ) : z ∈ D0 } − 2 for sufficiently small . By Theorem 6.3.12.2, ρ(T2P \ D0 ) > 1. Thus the potential  Gρ (z, ζ, D)ν(dζ) Π(z) = − D

exists and ν can be chosen in such way that supp ν  T2P \ D0 . By Proposition 6.2.3.6, ∂q0 > 0, z ∈ D0 . ∂n Thus ν can be chosen in such a way that −

∂q0 ∂Π < min , z ∈ ∂D0 . ∂n ∂n 

Then the function q(z) =

q0 (z), Π(z),

z ∈ D0 , z ∈ T2P \ D0 ,

is an L1 -subfunction on T2P . Exercise 6.3.12.2 Explain this in detail, exploiting Theorem 2.7.2.1. The function q(z) satisfies the condition q(z) ≤ m(z, GΛ , Λ) − 2 , ∀ z ∈ T2P , because of negative potential. Hence, q1 (z) := q(z) + η for some η > 0 also is a minorant of m(z, GΛ , Λ) and it is not minimal. Necessity is proved. Sufficiency follows from Theorem 6.3.7.1. 

6.3. Completeness of exponential systems in convex domains

245

6.3.13 Now we pass to the proof of Theorem 6.3.7.4 and construction of Example 6.3.7.2. Proof of Theorem 6.3.7.4. The set T2P \ D0 is closed. Let φ(z) be an infinitely differentiable function equal to zero on T2P \ D0 and positive on D0 . Set q(z) := h0 (y) − φ(z), where h0 (y) is a t.c.f., corresponding to G0 , and let be small enough to satisfy L1 q(z) > 0, z ∈ T2P . It is possible, because L1 h0 (y) > 0 by the condition of the theorem. Then we have m(z, G0 , q) = φ(z),

hence

{z : m(z, G0 , q) > 0} = D0 .

Take v(λ) := |λ|q(log λ) and construct an entire function ΦΛ for which Fr[ΦΛ ] = {v[t] : 1 ≤ t ≤ eP }. It is easy to check that the zero distribution of this function has all the properties demanded by Theorem 6.3.7.4.  Exercise 6.3.13.1 Check this. Proof of (6.3.7.8). Consider the problem L1 q(z) = 0, q|x=(2π/P )y = 0. Let us pass in the equation to new coordinates  ξ = x cos α + y sin α, η = −x sin α + y cos α, tan α = 2π/P. Then the equation takes the form: 

2  ∂ ∂ ∂2 ∂ − sin α + + 2ρ cos α r2 R(ξ, η) = 0. ∂ξ 2 ∂η 2 ∂ξ ∂η The condition of being zero on D0 is R(ξ, 2πl cos α) = 0. l ∈ Z. The condition of periodicity gives R(ξ + (P/ cos α)k, η) = R1 (ξ, η), k ∈ Z.

(6.3.13.1)

246

Chapter 6. Application to the Completeness of Exponential Systems

We search for a solution that does not depend on ξ. We have R (η) − 2ρ sin αR (η) + ρ2 R(η) = 0, R(0) = R(2π cos α) = 0. Further, R(η) = C1 e(ρ sin α)η cos((ρ cos α)η) + C1 e(ρ sin α)η sin((ρ cos α)η). Exploiting the boundary condition, we have −1

2

ρmin = (2 cos α)

1 1+ = 2



2π P

2 ! .

The corresponding eigenfunction is R = exp (ρmin sin α)η sin((ρmin cos α)η). It is zero on T2P \ D0 and positive in D0 , so it is determined up to a constant multiple.  6.3.14 We are going to prove Theorem 6.3.12.3. Actually we prove Theorem 6.3.14.1 Let Γ1 , Γ2 , . . . , Γn be Jordan curves, such that 1. Γi , i = 1, 2, . . . , n connect 0 and ∞; 2. there exists a number T, |T | > 1 (not necessarily real) for which T Γi = Γi , i = 1, 2, . . . , n. Let Di , i = 1, 2, . . . , n be domains into which the plane is divided, and let ρi be the order of the minimal harmonic function in Di . Then  1/ρi ≤ 2 (6.3.14.1) i

and equality holds if and only if Γi are a logarithmic spirals (or rays, when T ∈ R+ ). Proof. Denote by Hi the minimal harmonic function in Di . Then Hi = φi where φi : Di → Π+ is a conformal map of Di to the upper half-plane, φ(0) = 0. + + are continued by isomorphism to C, and The maps gi := φi (T φ−1 i ) : Π → Π gi (0) = 0. Thus gi (z) = σi z, where σi > 1. Hence, φi (T z) = σi φi (z) or : Π+ → Di . T hi (z) = hi (σi z), hi := φ−1 i Now we exploit the following inequality from [Lev] n  i=1

1 2 log T 2 . ≤ ≤ log σi | log T |2 log T

(6.3.14.2)

The equalities in (6.3.14.2) are attained only when Γi are logarithmic spirals or rays.  Since ρi = log σi / log |T | (6.3.14.2) implies (6.3.14.1).

Notation 2.1 Rm , M (f, x, ε), f ∗ (x), C + (E), C − (E), χG , χF , ΓA , F A , K, M (f, K), Kn , Kmax . 2.2 σ(G), μ, G0 (μ), supp μ, M(G), μF (E), ν, Md , ν + , ν − , |ν|, supp φ, ∗

◦

→, E, E, σ(Rm1 × Rm2 ), Φ1 ⊗ Φ2 , μ1 ⊗ μ2 . D

(n)

2.3 ϕn → ϕ, α(t), αε (x), ψε (x), f, ϕ, δx , ϕ, δx , ϕ, μ, ϕ, αf, ϕ, f1 + f2 , ϕ,  ∂x∂ k f, ϕ, f (x), f |G1 , c os ρ(φ). 2.4 Δx0 , Em (x), θm , G(x, y, Ω), G(x, y, Ka,R ). 2.5 Π(x, μ, D), GN (x, y), ΠN (x, μ, D), Π(x, μ), Π(z, μ), capG (K, D), capm (K), capm (D), capm (E), capm (E), capl (K). 2.6 M(x, r, u), N (x, r, u), E  , D− , u (x), KR , M (r, u), μ(r, u), M(r, u), N (r, u), M (z). 2.7 u˜(x), μx (t), E(α, α , , μ), En,δ0 . 2.8 a(r), ρ[a], σ[a], ρ(r), σ[a, ρ(r)], V (r), L(r), δSH(Rm ), T (r, u), ρT [u], ¯ ¯ ρ(r)], σT [u], σT [u, ρ(r)], ρM [u], σM [u], σM [u, ρ(r)], ρ[μ], Δ[μ], Δ[μ, N (r, μ), ρN [μ], δM(Rm ). 2.9 H(z, cos γ, m), G(x, y, Rm ), Dk (x, y), H(z, cos γ, m, p), Gp (x, y, m), R Gp (z, ζ, 2), Π(x, μ, p), δSH(ρ), ΠR < (x, ν, ρ − 1), Π> (x, ν, ρ), δR (x, ν, ρ), ¯ δ [u, ρ], Ω[u, ρ(r)], T (r, λ, >), T (r, λ, <). δR (z, ν, ρ), δR (x, u, ρ), M (r, δ), Δ 3.1 Vt , Pt , SH(Rm , ρ, ρ(r)), SH(ρ(r)), ut (x), Fr[u, ρ(r), V• , Rm ], U [ρ, σ], U [ρ], v[t] , M(Rm , ρ(r)), μ ∈ M(ρ(r)), Fr[μ, ρ(r), V• , Rm ], Fr[μ], M[ρ, Δ], M[ρ], ν[t] . 3.2 h(x, u), h(x, u), lx0 , x0 (x), Tρ , GI (φ, ψ), ΠI (φ, ds), T Cρ , CoΩ . cl

3.3 Δ(G, μ), Δ(E, μ), Δ(K, μ), Δ(E, μ), CoΩ (I), Δ (E), Δcl (E), ΩG ( ), ΩK ( ). 4.1 T t , (T • , M ), d(•, •), Ω(T • ), C(m), Ω(m), A(m), Tt v.

248

Notation

4.2 U0 , β(x), b0 , k(s), R v(x), Str(δ), v(x|t), v(•, t) 4.3 w(•|t), w(•|•). 4.4 u, (u)t , Fr[u], U [ρ]. 5.1 M (r, f ), T (r, f ), ρT [f ], ρM [f ], σT [f, ρ(r)], σM [f, ρ(r)], n(Kr ), n(r), ρ[n], Δ[n], N (r, n), ρN [n], ΔN [n], p[n], Fr[f ], Fr[n], Mer(ρ, ρ(r)), T (r, f ), ρT [f ], σT [f, ρ(r)]. 5.2 α − mesC, C0α , C00 . 5.3 g p . 5.4 h1 (φ, f ), h2 (φ, f ), h(φ, f ).   5.5 N (δ, X ), (X ) f dδ, (X ) E f dδ, δ(ΘF ), Dr,Θ , δz (Dr,Θ ), Acl (δ, χΘ ). 5.7 F (u), Hφ (u), T (u), Mα (u), M (u), Iαβ (u), I(u, g), F [f ], F [f ], χH , χI , χ F o . ˆ S1 − S), (Fν)(s). 5.8 KS1 , S1 , G(t, γ, ρ), G(s, ˆind . 6.1 H(z), m(z, v, H), GH , DH , Uind , U 6.2 T2P , D (T2P ), q(z), Lρ , Eρ (• − ζ), Eρ (• − ζ), qD , Gρ (z, ζ, D), Hρ (q). 6.3 Λ, ΦΛ (λ), exp Λ, A(G), hΛ (φ), GΛ , αG1 + βG2 , ΘΛ , IΛ , dΛ , hG (φ), m(λ, G, v), H(λ), qΛ (z), D(G, Λ), ρ(Λ, G), gG q, GG , DG , MIN, JG (Λ), HARM, m(φ, G, h), E(φ).

List of Terms 2.1 upper semicontinuous regularization upper semicontinuous function lower semicontinuous function 2.2 measure mass distribution support of μ μ is concentrated on E ∈ σ(G) restriction of μ onto F ∈ σ(G) charge positive and negative, respectively, variations of ν full variation of ν variation Borel function restriction of μ on the set E product of measures 2.3 linear space topological space linear continuous functional on D Schwartz distribution Dirac delta-function the nth derivative of the Dirac delta-function regular distribution positive distribution product of a distribution f by an infinitely differentiable function α(x) sum of distributions f1 and f2 partial derivative of distribution sequence of distributions fn converges to a distribution f regularization of the distribution f restriction of distribution f ∈ D (G) to G1 ⊂ G

250

List of Terms

fundamental solution of L at the point y spherical operator 2.4 harmonic distribution Lipschitz boundary, Lipschitz domain harmonic measure spherical function of a degree ρ Green potential of μ relative to D Newton potential logarithmic potential 2.5 balayage, sweeping Green capacity of the compact set K relative to the domain D Wiener capacity external and inner capacity of any set E capacible set logarithmic capacity irregular point equilibrium mass distribution h-Hausdorff measure Carleson measure 2.6 mean value of u(x) on the sphere Sx,r := {y : |y − x| = r} subharmonic function the least harmonic majorant of u in K Riesz measure of the subharmonic function u 2.7 precompact family of functions a sequence fn of locally summable functions converges in Lloc quasi-everywhere convergence a sequence of functions un converges to a function u relative to α-Carleson measure a point x ∈ Rm (α, α , )-normal with respect to the measure μ 2.8 order of a(r) type number of a(r) a(r) of minimal type a(r) of normal type a(r) of maximal type convergence exponent for the sequence {rj } a proximate order with respect to order ρ equivalent proximate orders type number with respect to a proximate order proper proximate order

List of Terms

Nevanlinna characteristic order of u(x) with respect to T (r) characteristics ρM [u], σM [u], σM [u, ρ(r)] convergence exponent of μ upper density of μ genus of μ N-order of μ N-type of μ 2.9 Gegenbauer polynomials Chebyshev polynomials primary kernel canonical potential zero distribution canonical Weierstrass product 3.1 limit set of the function u(x) limit set of the mass distribution μ 3.2 indicator of growth of u lower indicator ρ-subspherical function ρ-trigonometrically convex (ρ-t.c.) fundamental relation of indicator 3.3 upper (lower) density of μ subadditivity of Δ(E, •) superadditivity of Δ(E, •) semi-additivity generalized semi-additivity monotonic function of E ∈ Rm t-extension of E to be dense in angular densities 4.1 dynamical system ( , s)-chain from m to m chain recurrent dynamical system non-wandering point attractor completely regular growth polygonally connected set periodic dynamical system

251

252

4.2 partition of unit 4.3 pseudo-trajectory asymptotically dynamical pseudo-trajectory with dynamical asymptotics T• (a.d.p.t.) piecewise continuous pseudo-trajectory w(•|•) ω-dense pseudo-trajectory 4.4 subharmonic curve 5.1 entire function of order ρ and normal type with respect to proximate order ρ(r) entire function with prescribed limit set meromorphic function of order ρ and normal type with respect to a proximate order ρ(r) 5.2 relative Carleson α-measure 5.3 lower indicator of entire function 5.4 maximal interval of ρ-trigonometricity strictly ρ-t.c.f. concordant h and g 5.5 upper density of zeros of entire function (X )-integral with respect to a nonnegative measure δ 5.6 completely regular growth function CRG-function regular zero distribution regular zero distribution with integer ρ completely regular growth functions along curves of regular rotation curve of regular rotation 5.7 growth characteristic continuity, positive homogeneity asymptotic characteristics of growth total family of growth characteristics non-rarefied set rarefied set thinly closed set independent family of characteristic

List of Terms

List of Terms

6.1 ideally complementing H-multiplicator entire function is of minimal type with respect to a proximate order ρ(r), ρ(r) → ρ limit set of indicators the maximum principle for U [ρ] is valid in the domain G 6.2 automorphic connected on spirals spectrum strictly monotonic minimal v ∈ U [ρ] 6.3 function of exponential type completeness maximality extremal overcompleteness maximal domain of completeness extremely overcomplete system exp Λ trigonometrically convex function (t.c.f) conjugate indicator diagram regular set GΛ is enclosed in G enclosed with sliding enclosed hardly enclosed freely indicator limit set indicator set zero with tangency Λ is periodic w ∈ U [1] is minimal U ⊂ U [1] is minimal

253

Bibliography [An]

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Oberwolfach Seminars (OWS) The workshops organized by the Mathematisches Forschungsinstitut Oberwolfach are intended to introduce students and young mathematicians to current fields of research. By means of these well-organized seminars, also scientists from other fields will be introduced to new mathematical ideas. The publication of these workshops in the series Oberwolfach Seminars (formerly DMV seminar ) makes the material available to an even larger audience. ¨ OWS 38: Bobenko, A.I. / Schroder, P. / Sullivan, J.M. / Ziegler, G.M. (Eds.), Discrete Differential Geometry (2008). ISBN 978-3-7643-8620-7 Discrete differential geometry is an active mathematical terrain where differential geometry and discrete geometry meet and interact. It provides discrete equivalents of the geometric notions and methods of differential geometry, such as notions of curvature and integrability for polyhedral surfaces. Current progress in this field is to a large extent stimulated by its relevance for computer graphics and mathematical physics. This collection of essays, which documents the main lectures of the 2004 Oberwolfach Seminar on the topic, as well as a number of additional contributions by key participants, gives a lively, multi-facetted introduction to this emerging field. OWS 37: Galdi, G.P. / Rannacher, R. / Robertson, A.M. / Turek, S., Hemodynamical Flows (2008). ISBN 978-3-7643-7805-9 OWS 36: Cuntz, J. / Meyer, R. / Rosenberg, J.M., Topological and Bivariant K-theory (2007). ISBN 978-3-7643-8398-5 Topological K-theory is one of the most important invariants for noncommutative algebras. Bott periodicity, homotopy invariance, and various long exact sequences distinguish it from algebraic K-theory. We describe a bivariant K-theory for bornological algebras, which provides a vast generalization of topological K-theory. In addition, we discuss other approaches to bivariant K-theories for operator algebras. As applications, we study K-theory of crossed products, the Baum-Connes assembly map, twisted K-theory with some of its applications, and some variants of the Atiyah-Singer Index Theorem. OWS 35: Itenberg, I. / Mikhalkin, G. / Shustin, E., Tropical Algebraic Geometry (2007). ISBN 978-3-7643-8309-1 Tropical geometry is algebraic geometry over the semifield of tropical numbers, i.e., the real

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¨ Birkhauser Advanced Texts (BAT) Edited by Herbert Amann, Zurich ¨ University, Switzerland Steven G. Krantz, Washington University, St. Louis, USA Shrawan Kumar, University of North Carolina, Chapel Hill, USA ´ r, Universite´ Pierre et Marie Curie, Paris, France Jan Nekov aˇ This series presents, at an advanced level, introductions to some of the fields of current interest in mathematics. Starting with basic concepts, fundamental results and techniques are covered, and important applications and new developments discussed. The textbooks are suitable as an introduction for students and non–specialists, and they can also be used as background material for advanced courses and seminars.

Azarin, V. Growth Theory of Subharmonic Functions (2008). ISBN 978-3-7643-8885-0 Quittner, P. / Souplet, P. Superlinear Parabolic Problems. Blow-up, Global Existence and Steady States (2007). This book is devoted to the qualitative study of solutions of superlinear elliptic and parabolic partial differential equations and systems. This class of problems contains, in particular, a number of reaction-diffusion systems which arise in various mathematical models, especially in chemistry, physics and biology. The book is self-contained and up-to-date, it has a high didactic quality. It is devoted to problems that are intensively studied but have not been treated so far in depth in the book literature. The intended audience includes graduate and postgraduate students and researchers working in the field of partial differential equations and applied mathematics. ISBN 978-3-7643-8441-8 ´ Drabek, P. / Milota, J. Methods of Nonlinear Analysis. Applications to Differential Equations (2007). In this book, fundamental methods of nonlinear analysis are introduced, discussed and illustrated in straightforward examples. Every method considered is motivated and explained in its general form, but presented in an abstract framework as comprehensively as possible. Applications and generalizations are shown. In particular, a large number of methods is applied to boundary value problems for partial differential equations.

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